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>
#include <sys/types.h>
#include <unistd.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

const pid_t pid = getpid();
// Problem Specific Parameter:
#define error(args...) { vector<string> _debug = split(#args, ',');err(begin(_debug), args);}

vector<string> split(const string& s, char c){
	vector<string> v;stringstream ss(s);string x;
	while (getline(ss, x, c)) v.emplace_back(x);
	return move(v);
}

void err(vector<string>::iterator it) {cerr << endl;}
template<typename T, typename... Args> void err(vector<string>::iterator it, T a,Args... args){
	cerr << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << " ",err(++it, args...);
}

ll n, mod;
ll power(ll a, ll idx){
	ll b = 1LL;
	while(idx){
		if(idx & 1) b = b * a % mod;
		a = a * a % mod;
		idx >>= 1;
	}
	return b;
}

const int limit = 2010;
ll C[limit][limit];
ll P[limit][limit];

void init(){
	rep(i,limit)rep(j,i+1){
		if(j == 0 or j == i)
			C[i][j] = 1LL;
		else
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
	}
}


int main(void){
	cin >> n >> mod;
	init();
	
	ll ans = 0LL;	
	rep(i, n + 1){
		const int ni = n - i;
		ll cur = 0LL;
		rep(j, i + 1){
			const ll tmp = (power(2LL, j) + mod - 1) % mod;
			const ll unit = power(tmp, ni) % mod;
			const ll val = C[i][j] * unit % mod;
			if((i - j) % 2 == 1)
				cur = (cur + mod - val) % mod;
			else
				cur = (cur + val) % mod;	
		}

		cur = cur * C[n][i] % mod;
		cur = cur * power(2LL, i * (i - 1) / 2 + (n - i) * (n - i - 1) / 2) % mod;
		ans = (ans + cur) % mod;
	}

	cout << ans << endl;
	
	
	return 0;
}
