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<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_set>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
#include<tuple>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=1000000007;

ll mod;

pll Ex_gcd(ll a,ll b){
    if(b==0){return {1,0};}
    pll ret=Ex_gcd(b,a%b);
    ret.F-=a/b*ret.S;
    return {ret.S,ret.F};
}

ll R(ll a){
    ll ret=Ex_gcd(a,mod).F;
    ret%=mod;
    if(ret<0){ret+=mod;}
    return ret;
}

ll mod_pow(ll a,ll x){
    if(x==0){return 1;}
    if(a==0){return 0;}
    ll ret=1;
    while(x>0){
        if(x&1){(ret*=a)%=mod;}
        x>>=1;
        (a*=a)%=mod;
    }
    return ret;
}

class comb{
public:
    ll mod;
    ll mx;
    vector<ll> F;
    vector<ll> FR;
    

    comb(ll mod=1000000007,ll mx=100000):mod(mod),mx(mx),F(mx+1,1),FR(mx+1,1){
        mk_F();
    }
    
    ll mod_pow(ll a,ll x){
        a%=mod;
        ll ans=1;
        while(x>0){
            if(x&1){ans*=a; ans%=mod;}
            a*=a;
            a%=mod;
            x>>=1;
        }
        return ans;
    }
    
    pll Ex_gcd(ll a,ll b){
        if(b==0){return {1,0};}
        pll ret=Ex_gcd(b,a%b);
        ret.F-=a/b*ret.S;
        return {ret.S,ret.F};
    }
    
    ll prime_R(ll a){
        return mod_pow(a,mod-2);
    }
    
    ll R(ll a){
        ll ret=Ex_gcd(a,mod).F;
        ret%=mod;
        if(ret<0){ret+=mod;}
        return ret;
    }
    
    void mk_F(){
        for(ll i=1;i<=mx;i++){F[i]=F[i-1]*i%mod; FR[i]=R(F[i]);}
    }
    
    ll c(ll n,ll k){
        if(n<k){return 0;}
        if(n==k || k==0){return 1;}
        return F[n]*FR[n-k]%mod*FR[k]%mod;
    }
    
    //mod must be prime
    ll Lucas_C(ll n,ll m){
        ll ret=1;
        while(n>0 || m>0){
            ret*=c(n%mod,m%mod);
            ret%=mod;
            n/=mod; m/=mod;
        }
        return ret;
    }
    
    ll Stirling(ll n,ll k){
        ll ret=0;
        for(ll i=1;i<=k;i++){
            if((k-i)%2){ret-=c(k,i)*mod_pow(i,n)%mod;}
            else{ret+=c(k,i)*mod_pow(i,n)%mod;}
            ret%=mod;
        }
        ret*=R(F[k]);
        ret%=mod;
        if(ret<0){ret+=mod;}
        return ret;
    }
    
    ll Bell(ll n,ll k){
        ll ret=0;
        for(ll i=1;i<=k;i++){ret+=Stirling(n,i); ret%=mod;}
        return ret;
    }
};


int main(){
    ll n;
    cin>>n>>mod;
    comb C(mod,n);
    ll ans=0;
    for(int i=1;i<n;i++){
        ll cnt=0;
        for(int t=0;t<=i;t++){
            if(t&1){cnt-=mod_pow(mod_pow(2,i-t)-1,n-i)*C.c(i,t)%mod;}
            else{cnt+=mod_pow(mod_pow(2,i-t)-1,n-i)*C.c(i,t)%mod;}
        }
        cnt%=mod;
        ans+=cnt*C.c(n,i)%mod*mod_pow(2,i*(i-1)/2)%mod*mod_pow(2,(n-i)*(n-i-1)/2)%mod;
    }
    ans%=mod;
    if(ans<0){ans+=mod;}
    cout<<ans<<endl;
    
    
    
    return 0;
}
