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;

using ll = long long int;

ll N, M;

ll comb[2005][2005];

ll power(ll x, ll n){
    x %= M;
    ll res = 1;
    while(n > 0){
        if(n&1) res = res*x%M;
        x = x*x %M;
        n >>= 1;
    }
    return res;
}

int main(){
    cin >> N >> M;
    
    for(int i=0;i<=N;i++){
        for(int j=0;j<=i;j++){
            if(j==0 || j==i){
                comb[i][j] = 1;
            }else{
                comb[i][j] = (comb[i-1][j]+comb[i-1][j-1])%M;
            }
        }
    }

    ll ans = 0;
    for(ll a=1;a<N;a++){
        ll b = N-a;
        ll s = 0;
        for(ll x=0;x<=a;x++){
            s += (power(power(2,(a-x))-1+M,b)*comb[a][x]%M*(x&1?-1:1)+M)%M;
            s %= M;
        }
        ans += s*comb[N][a]%M*power(2,a*(a-1)/2)%M*power(2,b*(b-1)/2);
        ans %= M;
    }
    cout << ans << endl;
    return 0;
}
