Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

A sequence a is a subsegment of a sequence b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. We will denote the subsegments as [l, r], where l, r are two integers with 1 ≤ l ≤ r ≤ n. This indicates the subsegment where l-1 elements from the beginning and n-r elements from the end are deleted from the sequence.

For a permutation p_1, p_2, …, p_n, we define a framed segment as a subsegment [l,r] where max\\{p_l, p_{l+1}, ..., p_r\} - min\\{p_l, p_{l+1}, ..., p_r\} = r - l. For example, for the permutation (6, 7, 1, 8, 5, 3, 2, 4) some of its framed segments are: [1, 2], [5, 8], [6, 7], [3, 3], [8, 8]. In particular, a subsegment [i,i] is always a framed segments for any i between 1 and n, inclusive.

We define the happiness of a permutation p as the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n, and [l, r] is a framed segment. For example, the permutation [3, 1, 2] has happiness 5: all segments except [1, 2] are framed segments.

Given integers n and m, Jongwon wants to compute the sum of happiness for all permutations of length n, modulo the prime number m. Note that there exist n! (factorial of n) different permutations of length n.

Input

The only line contains two integers n and m (1 ≤ n ≤ 250 000, 10^8 ≤ m ≤ 10^9, m is prime).

Output

Print r (0 ≤ r < m), the sum of happiness for all permutations of length n, modulo a prime number m.

Examples

Input


1 993244853


Output


1


Input


2 993244853


Output


6


Input


3 993244853


Output


32


Input


2019 993244853


Output


923958830


Input


2020 437122297


Output


265955509

Note

For sample input n=3, let's consider all permutations of length 3:

  * [1, 2, 3], all subsegments are framed segment. Happiness is 6. 
  * [1, 3, 2], all subsegments except [1, 2] are framed segment. Happiness is 5. 
  * [2, 1, 3], all subsegments except [2, 3] are framed segment. Happiness is 5. 
  * [2, 3, 1], all subsegments except [2, 3] are framed segment. Happiness is 5. 
  * [3, 1, 2], all subsegments except [1, 2] are framed segment. Happiness is 5. 
  * [3, 2, 1], all subsegments are framed segment. Happiness is 6. 



Thus, the sum of happiness is 6+5+5+5+5+6 = 32.

import java.util.*;
import java.io.*;

public class C {
	static final int MOD_PRIME = 1000000007;

	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();

		int i = 0;
		int t = 1;
		//t = in.nextInt();
		for (; i < t; i++)
			solver.solve(i, in, out);

		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			long m = in.nextLong();
			long[] fact = new long[n+1];
			fact[0] = 1;
			for(int i = 1; i <= n; i++) {
				fact[i] = (fact[i-1]*i) % m;
			}
			long total = 0;
			for(int i = 0; i < n; i++) {
				total = (total + (n-i)*((fact[i+1] * fact[n-i]) % m)) % m;
			}
			out.print(total);
			
		}
	}

	// template code

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

	}

	static long modexp(long a, long b, long p) {
		// returns a to the power b mod p by modular exponentiation

		long res = 1;
		long mult = a % p;
		while (b > 0) {
			if (b % 2 == 1) {
				res = (res * mult) % p;
			}
			b /= 2;
			mult = (mult * mult) % p;
		}

		return res;
	}

	static double log(double arg, double base) {
		// returns log of a base b, contains floating point errors, dont use for exact
		// calculations.

		if (base < 0 || base == 1) {
			throw new ArithmeticException("base cant be 1 or negative");
		}
		if (arg < 0) {
			throw new ArithmeticException("log of negative number undefined");
		}
		return Math.log10(arg) / Math.log10(base);
	}

	static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}

		return gcd(b, a % b);

	}

	// scope for improvement
	static ArrayList<Integer> sieveOfEratosthenes(int n) {
		boolean[] check = new boolean[n + 1];
		ArrayList<Integer> prime = new ArrayList<Integer>();
		for (long i = 2; i <= n; i++) {
			if (!check[(int) i]) {
				prime.add((int) i);
				for (long j = i * i; j <= n; j += i) {
					check[(int) j] = true;
				}
			}
		}

		return prime;
	}

	static int modInverse(int a, int n) {
		// returns inverse of a mod n by extended euclidean algorithm
		int t = 0;
		int newt = 1;
		int r = n;
		int newr = a;
		int quotient;
		int tempr, tempt;
		while (newr != 0) {
			quotient = r / newr;
			tempt = newt;
			tempr = newr;
			newr = r - quotient * tempr;
			newt = t - quotient * tempt;
			t = tempt;
			r = tempr;
		}
		if (r > 1) {
			throw new ArithmeticException("inverse of " + a + " mod " + n + " does not exist");
		} else {
			if (t < 0) {
				t += n;
			}
			return t;
		}
	}

	static int primeModInverse(int a, int n) {
		// returns inverse of a mod n by mod exponentiation, use only if n is prime
		return (int) modexp(a, n - 2, n);
	}

	static void dfs(HashMap<Integer, ArrayList<Integer>> adj, Set<Integer> ans, Set<Integer> open,
			HashMap<String, Integer> edge, boolean[] vis, int i) {
		vis[i] = true;
		open.add(i);
		if (adj.get(i) != null) {
			for (int k : adj.get(i)) {
				if (!vis[k]) {
					dfs(adj, ans, open, edge, vis, k);
				} else if (open.contains(k)) {
					ans.add(edge.get(String.valueOf(i) + " " + String.valueOf(k)));
				}
			}
		}
		open.remove(i);
	}
}
