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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
//const ll mod=1e9+7;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};

const int M=2005;
ll n,mod,C[M][M];

ll Pow(ll x,ll p){
	ll r=1;
	for(;p>0;p>>=1){
		if(p&1) r=(r*x)%mod;
		x=(x*x)%mod;
	}
	return r;
}

int main(){
	cin>>n>>mod;
	C[0][0]++;
	for(int i=0;i<M-1;i++) for(int j=0;j<=i;j++){
		(C[i+1][j]+=C[i][j])%=mod;
		(C[i+1][j+1]+=C[i][j])%=mod;
	}
	ll res=0;
	for(ll i=1;i<n;i++){
		ll j=n-i,tmp=0;
		for(ll k=0;k<=j;k++){
			ll t=(k%2==0?1:mod-1);
			(t*=C[j][k])%=mod;
			(t*=Pow((Pow(2,j-k)-1+mod)%mod,i))%=mod;
			(tmp+=t)%=mod;
		}
		(tmp*=C[n][i])%=mod;
		(tmp*=Pow(2,i*(i-1)/2+j*(j-1)/2))%=mod;
		(res+=tmp)%=mod;
	}
	cout<<res<<endl;
}
