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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
ll mod;


vector<ll> inv,fact,invfact;
void mod_build(int n=101010){
	fact.resize(n+1);
	inv.resize(n+1);
	invfact.resize(n+1);
	fact[0]=inv[0]=invfact[0]=1;
	inv[1]=1;
	rep(i,n){
		fact[i+1]=fact[i]*(i+1)%mod;
		if(i>0)inv[i+1]=mod-inv[mod%(i+1)]*(mod/(i+1))%mod;
		invfact[i+1]=invfact[i]*inv[i+1]%mod;
	}
}
ll perm(int n,int k){
	if(n<0||k<0||k>n)return 0;
	return fact[n]*invfact[n-k]%mod;
}
ll comb(int n,int k){
	if(n<0||k<0||k>n)return 0;
	return (fact[n]*invfact[n-k]%mod)*invfact[k]%mod;
}
ll powmod(ll n,ll k){
	ll ret=1;
	while(k){
		if(k&1)ret=ret*n%mod;
		n=n*n%mod;
		k>>=1;
	}
	return ret;
}
int main(){
	ll n;
	cin>>n>>mod;
	mod_build(101010);
	ll ans=0;
	for(int i=1;i<=n-1;++i){
		ll res=0;
		for(int j=0;j<=i;++j){
			if(j%2)res+=comb(i,j)*(mod-powmod(powmod(2,(i-j))-1,n-i))%mod;
			else res+=comb(i,j)*powmod(powmod(2,(i-j))-1,n-i)%mod;
		}
		res%=mod;
		res=res*comb(n,i)%mod;
		ans+=res*powmod(2,(n-i)*(n-i-1)/2)%mod*powmod(2,i*(i-1)/2)%mod;
	}
	cout<<ans%mod<<endl;
	return 0;
}
