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>
typedef long long ll;
typedef long double ld;
const int INF=1e9,MOD=1e9+7;
const ll LINF=1e18;
using namespace std;
#define int long long
//template
int M;
int pw(int n,int k){
  int res=1;
  while(k){
    if(k%2)res*=n;res%=M;
    k/=2;
    n*=n;n%=M;
  }
  return res;
}

std::vector<int> v(1000000);
std::vector<int> w(1000000);
int nCk(int n,int k){
  int res=v[n];
  res*=w[k];res%=M;
  res*=w[n-k];res%=M;
  return res;
}

int frac(int N,int i){
  int res=0;
  for(int j=0;j<N-i;j++){
    int b=nCk(N-i,j)*pw(pw(2,N-i-j)-1,i)%M;
    if(j%2)b*=-1;
    res+=b;
  }
  res%=M;res+=M;res%=M;
  return res;
}
//main
signed main(){
  int N;cin>>N>>M;
  int ans=0;
  v[0]=1;for(int i=1;i<1000000;i++)v[i]=v[i-1]*i%M;
  for(int i=0;i<1000000;i++)w[i]=pw(v[i],M-2);
  for(int i=1;i<N;i++){
    int res=pw(2,i*(i-1)/2);
    res*=pw(2,(N-i)*(N-i-1)/2);res%=M;
    res*=frac(N,i);res%=M;
    res*=nCk(N,i);
    ans+=res;ans%=M;
    //cout<<pw(2,i*(i-1)/2)<<" "<<pw(2,(N-i)*(N-i-1)/2)<<" "<<frac(N,i)<<endl;
  }
  cout<<ans<<endl;
  //cout<<pw(3,5)<<" "<<nCk(7,3)<<endl;
}

