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 Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

Int MOD;
template<typename T>
struct Mint{
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }
  
  Mint inv(){return pow(MOD-2);}
  
  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}
  
  Mint operator+(Mint a) const{return Mint(v)+=a;};
  Mint operator-(Mint a) const{return Mint(v)-=a;};
  Mint operator*(Mint a) const{return Mint(v)*=a;};
  Mint operator/(Mint a) const{return Mint(v)/=a;};

  Mint operator-(){return v?MOD-v:v;}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static vector<Mint> fact,finv,invs;
  
  static void init(Int n){
    if(n+1<=(signed)fact.size()) return;
    fact.assign(n+1,1);
    finv.assign(n+1,1);
    invs.assign(n+1,1);
    
    for(Int i=1;i<=n;i++) fact[i]=fact[i-1]*Mint(i);
    finv[n]=Mint(1)/fact[n];
    for(Int i=n;i>=1;i--) finv[i-1]=finv[i]*Mint(i);
    for(Int i=1;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static Mint comb(long long n,Int k){
    Mint res(1);
    for(Int i=0;i<k;i++){
      res*=Mint(n-i);
      res/=Mint(i+1);
    }
    return res;
  }
  
  static Mint C(Int n,Int k){
    if(n<k||k<0) return Mint(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }

  static Mint P(Int n,Int k){
    if(n<k||k<0) return Mint(0);
    init(n);
    return fact[n]*finv[n-k];
  }
  
  static Mint H(Int n,Int k){
    if(n<0||k<0) return Mint(0);
    if(!n&&!k) return Mint(1);
    init(n+k-1);
    return C(n+k-1,k);
  }

  static Mint S(Int n,Int k){
    Mint res;
    init(k);
    for(Int i=1;i<=k;i++){
      Mint tmp=C(k,i)*Mint(i).pow(n);
      if((k-i)&1) res-=tmp;
      else res+=tmp;
    }    
    return res*=finv[k];
  }

  static vector<vector<Mint> > D(Int n,Int m){
    vector<vector<Mint> > dp(n+1,vector<Mint>(m+1,0));
    dp[0][0]=Mint(1);
    for(Int i=0;i<=n;i++){
      for(Int j=1;j<=m;j++){
        if(i-j>=0) dp[i][j]=dp[i][j-1]+dp[i-j][j];
        else dp[i][j]=dp[i][j-1];
      }
    }
    return dp;
  }

  static Mint B(Int n,Int k){
    Mint res;
    for(Int j=1;j<=k;j++) res+=S(n,j);
    return res;
  }

};
template<typename T>
vector<Mint<T> > Mint<T>::fact = vector<Mint<T> >();
template<typename T>
vector<Mint<T> > Mint<T>::finv = vector<Mint<T> >();
template<typename T>
vector<Mint<T> > Mint<T>::invs = vector<Mint<T> >();

//INSERT ABOVE HERE
const Int MAX = 2020;
using M = Mint<Int>;
M dp[MAX][MAX];
M sm[MAX][MAX];

signed main(){
  Int n;
  cin>>n>>MOD;
  M::init(MAX*2);
  
  vector<M> po(n+1,1);
  for(Int i=0;i<n;i++) po[i+1]=po[i]*M(2);
  vector<M> pp(n+1,0);
  for(Int i=0;i<n;i++) pp[i]=M(2).pow(i*(i-1)/2);
  
  M ans(0);
  for(Int i=1;i<n;i++){
    M res(0);
    for(Int j=0;j<=i;j++){
      M tmp=M::C(i,j);
      tmp*=(M(2).pow(j)-M(1)).pow(n-i);
      if(~(i-j)&1) res+=tmp;
      else res-=tmp;
    }
    ans+=res*M::C(n,i)*pp[i]*pp[n-i];
  }
  cout<<ans.v<<endl;
  return 0;
}

