Problem: Consider a permutation p_1, p_2, ... p_n of integers from 1 to n. We call a sub-segment p_l, p_{l+1}, ..., p_{r-1}, p_{r} of the permutation an interval if it is a reordering of some set of consecutive integers. For example, the permutation (6,7,1,8,5,3,2,4) has the intervals (6,7), (5,3,2,4), (3,2), and others. 

Each permutation has some trivial intervals — the full permutation itself and every single element. We call a permutation interval-free if it does not have non-trivial intervals. In other words, interval-free permutation does not have intervals of length between 2 and n - 1 inclusive. 

Your task is to count the number of interval-free permutations of length n modulo prime number p. 

Input

In the first line of the input there are two integers t (1 ≤ t ≤ 400) and p (10^8 ≤ p ≤ 10^9) — the number of test cases to solve and the prime modulo. In each of the next t lines there is one integer n (1 ≤ n ≤ 400) — the length of the permutation. 

Output

For each of t test cases print a single integer — the number of interval-free permutations modulo p. 

Examples

Input


4 998244353
1
4
5
9


Output


1
2
6
28146


Input


1 437122297
20


Output


67777575

Note

For n = 1 the only permutation is interval-free. For n = 4 two interval-free permutations are (2,4,1,3) and (3,1,4,2). For n = 5 — (2,4,1,5,3), (2,5,3,1,4), (3,1,5,2,4), (3,5,1,4,2), (4,1,3,5,2), and (4,2,5,1,3). We will not list all 28146 for n = 9, but for example (4,7,9,5,1,8,2,6,3), (2,4,6,1,9,7,3,8,5), (3,6,9,4,1,5,8,2,7), and (8,4,9,1,3,6,2,7,5) are interval-free.

The exact value for n = 20 is 264111424634864638. 

Solution (in JAVA):

Solution:
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.exit;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class I {

	static void solve() throws Exception {
		int tests = scanInt();
		int mod = scanInt();
		int facts[] = new int[401];
		facts[0] = 1;
		for (int i = 1; i <= 400; i++) {
			facts[i] = (int) ((long) facts[i - 1] * i % mod);
		}
		int factsInv[] = new int[401];
		factsInv[400] = BigInteger.valueOf(facts[400]).modInverse(BigInteger.valueOf(mod)).intValue();
		for (int i = 400; i > 0; i--) {
			factsInv[i - 1] = (int) ((long) factsInv[i] * i % mod);
		}
		int f2[] = new int[401];
		for (int i = 1; i <= 400; i++) {
			f2[i] = facts[i];
			for (int j = 1; j < i; j++) {
				f2[i] -= (long) f2[j] * facts[i - j] % mod;
				if (f2[i] < 0) {
					f2[i] += mod;
				}
			}
		}
		int cnt[][] = new int[401][401];
		cnt[0][0] = 1;
		int ans[] = new int[401];
		for (int i = 1; i <= 400; i++) {
			int cans = facts[i];
			for (int j = 1; j <= i; j++) {
				int v = 0;
				for (int k = 1; k <= i; k++) {
					v = (int) ((v + (long) cnt[i - k][j - 1] * facts[k]) % mod);
				}
				cnt[i][j] = v;
			}
			for (int j = 3; j < i; j++) {
				cans -= (long) ans[j] * cnt[i][j] % mod;
				if (cans < 0) {
					cans += mod;
				}
			}
			if (i != 2) {
				for (int j = 1; j < i; j++) {
					cans -= 2L * f2[j] * facts[i - j] % mod;
					if (cans < 0) {
						cans += mod;
					}
				}
			}
			ans[i] = cans;
		}
		for (int test = 0; test < tests; test++) {
			out.println(ans[scanInt()]);
		}
	}

	static int scanInt() throws IOException {
		return parseInt(scanString());
	}

	static long scanLong() throws IOException {
		return parseLong(scanString());
	}

	static String scanString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}