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<bits/stdc++.h>
using namespace std;

int main(){
    long N, M;
    cin >> N >> M;
    vector<vector<long>> C({{1}});
    for(int i = 1; i <= 2000; ++i){
        C.emplace_back(i + 1, 1);
        for(int j = 1; j < i; ++j)C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % M;
    }
    vector<long> b(2001);
    b[0] = 0;
    for(int i = 1; i <= 2000; ++i){
        b[i] = b[i - 1] * 2 + 1;
        b[i] -= (b[i] > M) * M;
    }
    auto modpow = [&M](long a, long n){
        long ret = 1;
        while(n){
            if(n & 1){
                ret *= a;
                ret %= M;
            }
            a *= a;
            a %= M;
            n >>= 1;
        }
        return ret;
    };
    long ans(0);
    for(long A = 0; A < N; ++A){
        long tmp(0), B(N - A);
        for(long i = 0; i <= B; ++i){
            long t(C[B][i] * modpow(b[B - i], A) % M);
            tmp += (i & 1) ? M - t : t;
        }
        ans += (tmp % M) * C[N][A] % M * modpow(2, A * (A - 1) / 2 + B * (B - 1) / 2) % M;
    }
    cout << ans % M << endl;
    return 0;
}
