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:

class ModInt {
public:
	static unsigned MOD;
	ModInt(): x(0) {}
	ModInt(signed y) : x(y >= 0 ? y % MOD : MOD - (-y) % MOD) {}
	ModInt(signed long long y) : x(y >= 0 ? y % MOD : MOD - (-y) % MOD) {}


	// Arithmetic Oprators
	ModInt &operator+=(ModInt that) {
		if ((x += that.x) >= MOD) x -= MOD; 
		return *this;
	}
	ModInt &operator-=(ModInt that) {
		if ((x += MOD - that.x) >= MOD) x -= MOD; 
		return *this;
	}
	ModInt &operator*=(ModInt that) {
		x = 1LL * x * that.x % MOD;
		return *this;
	}
	ModInt &operator/=(ModInt that) {
		return *this *= ModInt(get<1>(extgcd(that.x, int(MOD))));
	}
	ModInt &operator%=(ModInt that) {
		x %= that.x;
		return *this;
	}

	ModInt &operator+=(const int that) { return *this += ModInt(that);}
	ModInt &operator-=(const int that) { return *this -= ModInt(that);}
	ModInt &operator*=(const int that) { return *this *= ModInt(that);}
	ModInt &operator/=(const int that) { return *this /= ModInt(that);}
	ModInt &operator%=(const int that) { return *this %= ModInt(that);}

	// Comparators
	bool operator <(ModInt that) { return x < that.x; }
	bool operator >(ModInt that) { return x > that.x; }
	bool operator<=(ModInt that) { return x <= that.x; }
	bool operator>=(ModInt that) { return x >= that.x; }
	bool operator!=(ModInt that) { return x != that.x; }
	bool operator==(ModInt that) { return x == that.x; }

	// Utilities
	unsigned getval() const { return x;}
	ModInt operator+(ModInt that) const { return ModInt(*this) += that;}
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that;}
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that;}
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that;}
	ModInt operator%(ModInt that) const { return ModInt(*this) %= that;}
	ModInt operator+(const int that) const { return ModInt(*this) += that;}
	ModInt operator-(const int that) const { return ModInt(*this) -= that;}
	ModInt operator*(const int that) const { return ModInt(*this) *= that;}
	ModInt operator/(const int that) const { return ModInt(*this) /= that;}
	ModInt operator%(const int that) const { return ModInt(*this) %= that;}
	ModInt operator=(const int that) { return *this = ModInt(that);}
	friend istream &operator>>(istream& is, ModInt &that) {ll tmp; is >> tmp; that = ModInt(tmp);return is;}
	friend ostream &operator<<(ostream& os, const ModInt &that) { return os << that.x; }
	
	ModInt power(ll n) const {
		ll b = 1LL, a = x;
		while(n){
			if(n & 1) b = b * a % MOD;
			a = a * a % MOD;
			n >>= 1;
		}
		return ModInt(b);
	}
private:
	unsigned x;
	
	inline tuple<int, int, int> extgcd(int a, int b) {
		if (b == 0) return make_tuple(a, 1, 0);
		tuple<int, int, int> ret = extgcd(b, a % b);
		swap(get<1>(ret), get<2>(ret));
		get<2>(ret) -= a / b * get<1>(ret);
		return ret;
	}
};

unsigned ModInt::MOD = UINT_MAX;
using mint = ModInt;
const mint ZERO = mint(0);
const mint ONE = mint(1);
const mint TWO = mint(2);

vector<mint> Fact, InvFact;
void makeFact(int n){
	Fact = vector<mint>(n+1);
	Fact[0] = mint(1);
	rep(i,1,n+1) Fact[i] = mint(i) * Fact[i-1];

	InvFact = vector<mint>(n+1);
	InvFact[n] = mint(1) / Fact[n];
	rrep(i,n) InvFact[i] = mint(i+1) * InvFact[i+1];
}

mint Factorial(int n){ return Fact[n];}
mint InverseFactorial(int n){ return InvFact[n];}
mint Permutation(int n,int k){ return (n < 0 or k < 0 or n - k < 0) ? ZERO : Fact[n] * InvFact[n - k];}
mint Combination(int n,int k){ return (n < 0 or k < 0 or n - k < 0) ? ZERO : Fact[n] * InvFact[k] * InvFact[n-k];}


int main(void){
	int n;
	cin >> n >> ModInt::MOD;
	makeFact(n);

	mint ans = ZERO;
	rep(i, n + 1){
		const int ni = n - i;
		mint cur = ZERO;
		rep(j, i + 1){
			const mint tmp = TWO.power(j) - ONE;
			const mint unit = tmp.power(ni);
			const mint val = Combination(i, j) * unit;
			if((i - j) % 2 == 1)
				cur -= val;
			else
				cur += val;
		}

		cur *= Combination(n, i) * TWO.power(i * (i - 1) / 2 + (n - i) * (n - i - 1) / 2);
		ans += cur;
	}

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