Problem: In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.

Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:

* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.



You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.

Constraints

* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.

Examples

Input

2 1000000007


Output

2


Input

3 1000000009


Output

118


Input

50 111111113


Output

1456748


Input

3000 123456791


Output

16369789

Solution (in CPP):

Solution:
#include<cstdio>
using namespace std;
typedef long long ll;
const int N=3005;
int ans,n,mod,jc[N],inv[N],s[N][N],f[N],pw[N*N];
int ksm(int x,int y,int mod)
{
  int res=1;
  while (y) {if (y&1) res=(ll)res*x%mod;x=(ll)x*x%mod;y>>=1;}
  return res;
}
int c(int n,int m) {return (ll)jc[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
	scanf("%d%d",&n,&mod);
	jc[0]=jc[1]=inv[0]=inv[1]=pw[0]=1;
	for (int i=2;i<=n;i++) jc[i]=(ll)jc[i-1]*i%mod,inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod;
	for (int i=2;i<=n;i++) inv[i]=(ll)inv[i-1]*inv[i]%mod;
	for (int i=1;i<=n*n;i++) pw[i]=(ll)pw[i-1]*2%mod;
	for (int i=0;i<=n;i++) s[i][0]=1;//初始化，都有一种可能分到0个集合 
	for (int i=1;i<=n;i++)
	  for (int j=1;j<=i;j++)
	    s[i][j]=((ll)s[i-1][j-1]+(ll)(j+1)*s[i-1][j]%mod)%mod;//<=i个元素分入j个非空集合的方案数  
	for (int i=0;i<=n;i++)
	{
	   for (int j=0;j<=i;j++) f[i]=((ll)f[i]+(ll)s[i][j]*pw[(n-i)*j]%mod)%mod;
	   f[i]=(ll)f[i]*ksm(2,ksm(2,n-i,mod-1),mod)%mod;	
	}
	for (int i=0;i<=n;i++)
	  if (i&1) ans=((ll)ans-(ll)c(n,i)*f[i]%mod+mod)%mod;
	  else ans=((ll)ans+(ll)c(n,i)*f[i]%mod)%mod;
	printf("%d\n",ans);
	return 0;
}