Note that this is the second problem of the two similar problems. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.

You are given a tree with n nodes. In the beginning, 0 is written on all edges. In one operation, you can choose any 2 distinct leaves u, v and any integer number x and add x to values written on all edges on the simple path between u and v. Note that in previous subtask x was allowed to be any real, here it has to be integer.

For example, on the picture below you can see the result of applying two operations to the graph: adding 2 on the path from 7 to 6, and then adding -1 on the path from 4 to 5. 

<image>

You are given some configuration of nonnegative integer pairwise different even numbers, written on the edges. For a given configuration determine if it is possible to achieve it with these operations, and, if it is possible, output the sequence of operations that leads to the given configuration. Constraints on the operations are listed in the output format section.

Leave is a node of a tree of degree 1. Simple path is a path that doesn't contain any node twice.

Input

The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of nodes in a tree.

Each of the next n-1 lines contains three integers u, v, val (1 ≤ u, v ≤ n, u ≠ v, 0 ≤ val ≤ 10 000), meaning that there is an edge between nodes u and v with val written on it. It is guaranteed that these edges form a tree. It is guaranteed that all val numbers are pairwise different and even. 

Output

If there aren't any sequences of operations which lead to the given configuration, output "NO".

If it exists, output "YES" in the first line. In the second line output m — number of operations you are going to apply (0 ≤ m ≤ 10^5). Note that you don't have to minimize the number of the operations!

In the next m lines output the operations in the following format:

u, v, x (1 ≤ u, v ≤ n, u not = v, x — integer, -10^9 ≤ x ≤ 10^9), where u, v — leaves, x — number we are adding. 

It is guaranteed that if there exists a sequence of operations producing given configuration, then there exists a sequence of operations producing given configuration, satisfying all the conditions above.

Examples

Input


5
1 2 2
2 3 4
3 4 10
3 5 18


Output


NO

Input


6
1 2 6
1 3 8
1 4 12
2 5 2
2 6 4


Output


YES
4
3 6 1
4 6 3
3 4 7
4 5 2

Note

The configuration from the first sample is drawn below, and it is impossible to achieve.

<image>

The sequence of operations from the second sample is illustrated below.

<image>
import static java.lang.Math.*;

import java.util.*;
import java.util.stream.*;

public class A2 {

	int N, M = 2000; int [][][] G;
	int [] H, L;
	HashMap<Long, Integer> res = new HashMap<>();

	public Object solve () {
		N = sc.nextInt();
		int [][] E = dec(sc.nextInts(N-1), 2);

		G = wgraph(N, E);
		for (int i : rep(N))
			if (G[0][i].length == 2)
				return "NO";

		H = new int [N]; L = new int [N];
		for (int i : rep(N))
			if (G[0][i].length == 1) {
				dfs(i, -1, 0);
				break;
			}

		print("YES");
		print(res.size());
		for (long key : res.keySet()) {
			long x = key / M, y = key % M;
			print(x+1, y+1, res.get(key));
		}
		return null;
	}

	void dfs (int s, int p, int v) {
		int W = 0;
		for (int m : rep(G[0][s].length)) {
			int j = G[0][s][m], w = G[1][s][m];
			if (j != p) {
				dfs(j, s, w);
				L[s] = L[j];
				W += w;
			}
		}

		if (p == -1) {
			H[s] = G[1][s][0];
			for (int i : rep(N))
				if (i != s)
					addEdge(s, i, H[i]);
			for (int i : rep(N))
				assert H[i] == 0;
			return;

		}

		if (G[0][s].length == 1) {
			H[s] = v; L[s] = s;
			return;
		}

		if (W != v) {
			int n = 0;
			while (G[0][s][n] == p) ++n;
			int j = G[0][s][n++];
			while (G[0][s][n] == p) ++n;
			int k = G[0][s][n++];
			addEdge(L[j], L[k], (W - v) / 2);
		}
	}

	void addEdge(int j, int k, int v) {
		if (v == 0)
			return;
		int x = min(j, k), y = max(j, k);
		long key = M * x + y;
		int w = res.getOrDefault(key, 0) + v;
		res.remove(key);
		if (w != 0)
			res.put(key, w);
		H[j] -= v; H[k] -= v;
	}

	private static final int CONTEST_TYPE = 1;

	private static void init () {
	}

	private static int [][] dec (int [][] E, int N) { for (int [] e : E) for (int i = 0; i < e.length && i < N; ++i) --e[i]; return E; }
	private static int [][] dup (int [][] E) {
		int [][] res = new int [2*E.length][];
		for (int i = 0; i < E.length; ++i) {
			res[2*i] = E[i].clone();
			res[2*i+1] = E[i].clone();
			res[2*i+1][0] = E[i][1]; res[2*i+1][1] = E[i][0];
		}
		return res;
	}
	private static int [][][] dwgraph (int N, int [][] E) {
		int [] D = new int [N];
		for (int [] e : E)
			++D[e[0]];
		int [][][] res = new int [2][N][];
		for (int i = 0; i < 2; ++i)
			for (int j = 0; j < N; ++j)
				res[i][j] = new int [D[j]];
		D = new int [N];
		for (int [] e : E) {
			int x = e[0], y = e[1], z = e.length > 2 ? e[2] : 1;
			res[0][x][D[x]] = y;
			res[1][x][D[x]] = z;
			++D[x];
		}
		return res;
	}
	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	private static int [][][] wgraph (int N, int [][] E) { return dwgraph(N, dup(E)); }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static Object print (Object o, Object ... A) { IOUtils.print(o, A); return null; }
	private static class IOUtils {
		public static class MyScanner {
			public String next () { newLine(); return line[index++]; }
			public int nextInt () { return Integer.parseInt(next()); }
			public String nextLine () { line = null; return readLine(); }
			public String [] nextStrings () { return split(nextLine()); }
			public int[] nextInts () { return nextStream().mapToInt(Integer::parseInt).toArray(); }
			public int[][] nextInts (int N) { return IntStream.range(0, N).mapToObj(i -> nextInts()).toArray(int[][]::new); }
			//////////////////////////////////////////////
			private boolean eol () { return index == line.length; }
			private String readLine () {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine () {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private java.util.stream.Stream<String> nextStream () { return java.util.Arrays.stream(nextStrings()); }
			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim (String delim, Object o, Object ... A) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : A)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static final java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, final Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (final StringBuilder b, Object o, final String delim) {
			append(x -> { append(b, x, delim); }, x -> b.append(delim).append(x), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void print (Object o, Object ... A) {
			String res = build(o, A);
			if (DEBUG == 2)
				err(res, '(', time(), ')');
			pw.println(res);
			if (DEBUG == 1) {
				pw.flush();
				System.out.flush();
			}
		}
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static int DEBUG;
		private static void exit () {
			String end = "------------------" + System.lineSeparator() + time();
			switch(DEBUG) {
			case 1: print(end); break;
			case 2: err(end); break;
			}
			IOUtils.pw.close();
			System.out.flush();
			System.exit(0);
		}
		private static long t;
		private static long millis () { return System.currentTimeMillis(); }
		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
		private static void run (int N) {
			try { DEBUG = Integer.parseInt(System.getProperties().get("DEBUG").toString()); }
			catch (Throwable t) {}

			for (int n = 1; n <= N; ++n) {
				Object res = new A2().solve();
				if (res != null) {
					@SuppressWarnings("all")
					Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
					print(o);
				}
			}
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main (String[] args) {
		init();
		@SuppressWarnings("all")
		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
		IOUtils.run(N);
	}
}
