CodeForces - 622F：The Sum of the k-th Powers （拉格朗日插值法求自然数幂和）

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.

Examples

Input

4 1

Output

10

Input

4 2

Output

30

Input

4 3

Output

100

Input

4 0

Output

4

就是抄个板子在这里。

#include<bits/stdc++.h>
#define ll long long
const int maxn=1000005;
const int mod=1000000007;
using namespace std;
ll f[maxn],fac[maxn],inv[maxn];
ll P(ll a,ll b)
{
    ll ans=1;
    while(b) {
        if(b&1) ans=ans*a%mod;
        b>>=1; a=a*a%mod;
    }
    if(ans<0) ans+=mod;
    return ans;
}
void init(int tot)
{
    fac[0]=1;
    for(int i=1;i<=tot;i++)
        fac[i]=fac[i-1]*i%mod;
    inv[tot]=P(fac[tot],mod-2);
    inv[0]=1; //求阶乘逆元 
    for(int i=tot-1;i>=1;i--)
        inv[i]=inv[i+1]*(i+1)%mod;
}
ll Lagrange(ll n,ll k)
{
    int tot=k+1;
    init(tot);
    ll ans=0,now=1;
    for(int i=1;i<=tot;i++) now=now*(n-i)%mod;
    for(int i=1;i<=tot;i++) {
        ll inv1=P(n-i,mod-2);
        ll inv2=inv[i-1]*inv[tot-i]%mod;
        if((tot-i)&1) inv2=mod-inv2;
        ll temp=now*inv1%mod;
        temp=temp*f[i]%mod*inv2%mod;
        ans+=temp;
        if(ans>=mod) ans-=mod;
    }
    return ans;
}
int main()
{
    ll n,k;
    cin>>n>>k;
    for(int i=1;i<=k+2;i++) f[i]=(f[i-1]+P(i,k))%mod;
    if(n<=k+2) cout<<f[n]<<endl;
    else cout<<Lagrange(n,k+1)<<endl;
    return 0;
}