F: Red-Black Soul Gem

problem

Homu-chan has the mysterious power to change the soul game to red or black, and to connect two different soul games with a magic thread. By using this power, Homu-chan can create a magic square by Sorujemu.

Homu-chan was wondering how many different magic squares there were when there were N numbered magic squares numbered from 1 to N, and the following conditions were met:

* You can only connect once between any of the two sources.
* All the colors of the game are either red or black.
* For any game, it is connected to at least one of the games whose color is different from its own.



At this time, the magic square can be regarded as a graph with the sorujemu as the apex and the magic thread as the side. Note that the graph does not have to be concatenated.

Homu-chan is not good at calculating, so let's create a fast program instead and calculate the number of different magic squares. However, the number is expected to be very large, so we will find the remainder after dividing by a prime number M.

Note that the magic squares G and H are different because the colors of G and H are different for a certain magic square v, or the pair u and v of a certain magic square is one of G and H. It means that it is connected only with.

Input format


N M

Constraint

* 2 \ leq N \ leq 2,000
* 10 ^ 8 \ leq M \ leq 10 ^ 9 + 7
* M is guaranteed to be prime



Output format

Divide the number of graphs that satisfy the condition by M and output the remainder as an integer on one line.

Input example 1


3 310779401

Output example 1


12

For example, the following magic squares meet the conditions.

Input example 2


7 566666561

Output example 2


98638848

For example, the following magic squares meet the conditions.

Input example 3


1010 1000000007

Output example 3


862232855

Note that we output the remainder after dividing by M.





Example

Input

3 310779401


Output

12

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <cstring>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <array>
#include <cassert>
#include <bitset>
using namespace std;
using LL = long long;

LL N, M;
LL fact[123456];
LL finv[123456];

LL mpow(LL a, LL x) {
	if (x == 0)return 1;
	LL tmp = mpow(a, x / 2);
	tmp *= tmp;
	tmp %= M;
	if (x % 2)tmp *= a;
	tmp %= M;
	return tmp;
}

inline LL comb(LL n, LL r) {
	if (r < 0 || n < r)return 0;
	LL tmp = fact[n] * finv[r];
	tmp %= M;
	tmp *= finv[n - r];
	tmp %= M;
	return tmp;
}

int main(void)
{
	cin >> N >> M;
	fact[0] = finv[0] = 1;
	for (LL i = 1; i < 123400; ++i) {
		fact[i] = fact[i - 1] * i;
		fact[i] %= M;
		finv[i] = mpow(fact[i], M - 2);
	}
	LL total = 0;
	for (int p = 1; p < N; ++p) {
		int q = N - p;
		LL ans = comb(N, p);
		ans *= mpow(2, comb(p, 2));
		ans %= M;
		ans *= mpow(2, comb(q, 2));
		ans %= M;
		LL sum = 0;
		LL mone = 1;
		for (int j = 0; j <= p; ++j) {
			LL tmp = (mone * comb(p, j)) % M;
			tmp *= mpow((mpow(2, p - j) + M - 1) % M, q);
			tmp %= M;
			sum += tmp;
			sum %= M;
			mone *= (M - 1);
			mone %= M;
		}
		ans *= sum;
		ans %= M;
		total += ans;
		total %= M;
	}
	cout << total << endl;
	return 0;
}

