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 int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 1000000000000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;



#define SIZE 5000005


ll N,M;
ll POW[SIZE];
ll fact[SIZE],inv_fact[SIZE];

ll mod_pow(ll x,ll count, ll mod){

	if(count == 0)return 1;
	ll ret = mod_pow((x*x)%mod,count/2,mod);
	if(count%2 == 1){

		ret = (ret*x)%mod;
	}
	return ret;
}

ll extgcd(ll a,ll b,ll &x,ll &y){
	ll d = a;
	if(b != 0){
		d = extgcd(b,a%b,y,x);
		y -= (a/b)*x;
	}else{
		x = 1;
		y = 0;
	}
	return d;
}

ll mod_inverse(ll a,ll m){
    ll x,y;
    extgcd(a,m,x,y);
    return (m+x%m)%m;
}

ll nCk(ll n,ll k){

	ll ret = fact[n]*inv_fact[k];
	ret %= M;
	ret *= inv_fact[n-k];

	return ret%M;
}



int main(){

	scanf("%lld %lld",&N,&M);

	POW[0] = 1;
	for(int i = 1; i < SIZE; i++){

		POW[i] = POW[i-1]*2;
		POW[i] %= M;
	}

	fact[0] = 1;
	for(ll i = 1; i < SIZE; i++){
		fact[i] = i*fact[i-1];
		fact[i] %= M;
	}
	inv_fact[SIZE-1] = mod_inverse(fact[SIZE-1],M);
	for(ll i = SIZE-1; i >= 1; i--){

		inv_fact[i-1] = inv_fact[i]*i;
		inv_fact[i-1] %= M;
	}

	ll ans = 0;

	for(ll r = 1; r <= N-1; r++){

		ll tmp = nCk(N,r);
		ll b = N-r;

		tmp *= POW[(r*(r-1))/2];
		tmp %= M;
		tmp *= POW[(b*(b-1))/2];
		tmp %= M;

		ll mult = 0;

		for(ll n = 0; n <= b; n++){
			ll work = nCk(b,n);
			ll rest = b-n;

			ll num = (POW[rest]-1+M)%M;
			work *= mod_pow(num,r,M);

			work %= M;

			if(n%2 == 1){

				mult -= work;
				if(mult < 0){

					mult += M;
				}

			}else{

				mult += work;
				mult %= M;
			}
		}

		ans += (tmp*mult)%M;
		ans %= M;
	}

	printf("%lld\n",ans);

	return 0;
}

