One Big Software Company has n employees numbered from 1 to n. The director is assigned number 1. Every employee of the company except the director has exactly one immediate superior. The director, of course, doesn't have a superior.

We will call person a a subordinates of another person b, if either b is an immediate supervisor of a, or the immediate supervisor of a is a subordinate to person b. In particular, subordinates of the head are all other employees of the company.

To solve achieve an Important Goal we need to form a workgroup. Every person has some efficiency, expressed by a positive integer ai, where i is the person's number. The efficiency of the workgroup is defined as the total efficiency of all the people included in it.

The employees of the big software company are obsessed with modern ways of work process organization. Today pair programming is at the peak of popularity, so the workgroup should be formed with the following condition. Each person entering the workgroup should be able to sort all of his subordinates who are also in the workgroup into pairs. In other words, for each of the members of the workgroup the number of his subordinates within the workgroup should be even.

Your task is to determine the maximum possible efficiency of the workgroup formed at observing the given condition. Any person including the director of company can enter the workgroup.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — the number of workers of the Big Software Company. 

Then n lines follow, describing the company employees. The i-th line contains two integers pi, ai (1 ≤ ai ≤ 105) — the number of the person who is the i-th employee's immediate superior and i-th employee's efficiency. For the director p1 = - 1, for all other people the condition 1 ≤ pi < i is fulfilled.

Output

Print a single integer — the maximum possible efficiency of the workgroup.

Examples

Input

7
-1 3
1 2
1 1
1 4
4 5
4 3
5 2


Output

17

Note

In the sample test the most effective way is to make a workgroup from employees number 1, 2, 4, 5, 6.
//package vk2015.r2;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class B2 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni();
		int[] par = new int[n];
		int[] sc = new int[n];
		for(int i = 0;i < n;i++){
			par[i] = ni()-1;
			if(par[i] == -2)par[i] = -1;
			sc[i] = ni();
		}
		int[][] g = parentToG(par);
		int[][] pars = parents3(g, 0);
		int[] ord = pars[1];
		long[][] dp = new long[n][2];
		for(int i = n-1;i >= 0;i--){
			int cur = ord[i];
			long odd = Long.MIN_VALUE / 3, even = 0;
			for(int e : g[cur]){
				if(par[cur] != e){
					long nodd = Math.max(odd + dp[e][0], even + dp[e][1]);
					long neven = Math.max(odd + dp[e][1], even + dp[e][0]);
					odd = nodd; even = neven;
				}
			}
			odd = Math.max(odd, even + sc[cur]);
			dp[cur][0] = even;
			dp[cur][1] = odd;
		}
//		tr(dp);
		out.println(Math.max(dp[0][0], dp[0][1]));
	}
	
	public static int[][] parents3(int[][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);

		int[] depth = new int[n];
		depth[0] = 0;

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int nex : g[cur]) {
				if (par[cur] != nex) {
					q[r++] = nex;
					par[nex] = cur;
					depth[nex] = depth[cur] + 1;
				}
			}
		}
		return new int[][] { par, q, depth };
	}
	
	public static int[][] parentToG(int[] par)
	{
		int n = par.length;
		int[] ct = new int[n];
		for(int i = 0;i < n;i++){
			if(par[i] >= 0){
				ct[i]++;
				ct[par[i]]++;
			}
		}
		int[][] g = new int[n][];
		for(int i = 0;i < n;i++){
			g[i] = new int[ct[i]];
		}
		for(int i = 0;i < n;i++){
			if(par[i] >= 0){
				g[par[i]][--ct[par[i]]] = i;
				g[i][--ct[i]] = par[i];
			}
		}
		return g;
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new B2().run(); }
	
	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
