Discription

You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is x^k.

Output the sum of costs over all non-empty subsets of people.

Input

Only line of input contains two integers N (1?≤?N?≤?10⁹) representing total number of people and k

Output

Output the sum of costs for all non empty subsets modulo 10⁹?+?7.

Example

1 1

1

3 2

24

Note

In the first example, there is only one non-empty subset {1} with cost 1¹?=?1.

In the second example, there are seven non-empty subsets.

- {1} with cost 1²

- {2} with cost 1²?=?1

- {1,?2} with cost 2²?=?4

- {3} with cost 1²?=?1

- {1,?3} with cost 2²?=?4

- {2,?3} with cost 2²?=?4

- {1,?2,?3} with cost 3²?=?9

The total cost is 1?+?1?+?4?+?1?+?4?+?4?+?9?=?24.

題目大意就是要你求一下ΣC(n,i)*i^k。

然後直接上我推的式子了(就是用第二類斯特林數代換一下)

(怎麽這個圖這麽大。。。。不管了)

然後就開開心心A了。

不過好像k再大一點也可以做，，，就是要用FFT 在N log N的時間求出某一行的斯特林數了(反正我也不會hhhh)，不能再N^2遞推斯特林數了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define ll long long
#define maxn 5005
using namespace std;
const int ha=1000000007;
const int inv=ha/2+1;
int S[maxn],n,m;
int ans=0,tmp,ci;

inline void init(){
	S[1]=1;
	for(int i=2;i<=m;i++)
	    for(int j=i;j;j--){
	    	S[j]=(S[j]*(ll)j+(ll)S[j-1])%ha;
		}
}

inline int ksm(int x,int y){
	int an=1;
	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
	return an;
}

inline void solve(){
	tmp=1,ci=ksm(2,n);
	for(int i=1;i<=m;i++){
		tmp=tmp*(ll)(n-i+1)%ha;
		ci=ci*(ll)inv%ha;
		ans=((ll)ans+S[i]*(ll)tmp%ha*(ll)ci)%ha;
	}
}

int main(){
	scanf("%d%d",&n,&m);
	init();
	solve();
	printf("%d\n",ans);
	return 0;
}

Codeforces 932 E Team Work