Codeforces 622F The Sum of the k-th Powers

Discription

There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees.

Find the value of the sum  modulo 10⁹ + 7 (so you should find the remainder after dividing the answer by the value 10⁹ + 7).

Input

The only line contains two integers n, k (1 ≤ n ≤ 10⁹, 0 ≤ k ≤ 10⁶).

Output

Print the only integer a — the remainder after dividing the value of the sum by the value 10⁹ + 7.

Example

Input

4 1

Output

10

Input

4 2

Output

30

Input

4 3

Output

100

Input

4 0

Output

4

拉格朗日差值裸题。。。
为什么我以前都用高斯消元做自然幂数和2333333
拉格朗日差值的构造原理和中国剩余定理类似，这里就不在赘述，网上也有很多讲的。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000010;
const int ha=1000000007;
int jc[maxn+5],TT,l;
int n,k,ans=0,base;

inline int add(int x,int y){
	x+=y;
	return x>=ha?x-ha:x;
}

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 init(){
	jc[0]=1;
	for(int i=1;i<=maxn;i++) jc[i]=jc[i-1]*(ll)i%ha;
	l=k+2,base=1;
}

inline void solve(){
	if(n<=l){
		for(int i=1;i<=n;i++) ans=add(ans,ksm(i,k));
	}
	else{
		for(int i=1;i<=l;i++) base=base*(ll)(n-i)%ha;
		for(int i=1,now;i<=l;i++){
			TT=add(TT,ksm(i,k));
			now=base*(ll)ksm(n-i,ha-2)%ha*(ll)ksm(jc[i-1],ha-2)%ha*(ll)ksm(jc[l-i],ha-2)%ha;
			if((l-i)&1) now=now*(ll)(ha-1)%ha;
			ans=add(ans,TT*(ll)now%ha);
		}
	}
}

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