CF622F The Sum of the k-th Powers (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).

 1 #include"bits/stdc++.h"
 2 using namespace std;
 3 
 4 #define int long long
 5 
 6 int n,k;
 7 const int mod = 1e9+7;
 8 const int N = 2e6;
 9 int fac[N];
10 
11 int ksm(int a,int b)
12 {
13     int ans = 1;
14     a%=mod;
15     for(; b; b>>=1,a*=a,a%=mod)
16         if(b&1)ans*=a,ans%=mod;
17     return ans;
18 }
19 void lg()
20 {
21 
22     int ans = 0;
23     fac[1]=1; fac[0]=1;
24     for(int i=2; i<=k+10; i++)fac[i]=(fac[i-1]*i%mod);
25     /// 预处理阶乘
26     int now=0;
27     int fz=1;
28     for(int i=1; i<=k+2; i++)
29         fz*=(n-i),fz%=mod; /// 处理分子
30 
31     for(int i=1; i<=k+2; i++)
32     {
33         now += ksm(i,k);    /// 每次的yi
34         now%=mod;
35         int inv1=ksm(n-i,mod-2);///少的分子的逆元
36         int inv2=ksm(fac[i-1]*fac[k+2-i],mod-2);
37         ///分母的逆元，分母可以看成是两段阶乘的乘积
38         
39         int f=1;
40         if((k+2-i)%2==1)f=-1;
41         ans+=f*fz*inv1%mod*inv2%mod*now%mod;
42         ans%=mod;
43 
44      //   cout<<i<<" "<<ans<<endl;
45     }
46     cout<<(ans+mod)%mod;
47 
48 
49 }
50 
51 signed main()
52 {
53     //cout<<ksm(2,5);
54     cin>>n>>k;
55     if(k==0)
56     {
57         cout<<n;
58         return 0;
59     }
60     if(n<=k+2)
61     {
62         int now=0;
63         for(int i=1; i<=n; i++)
64             now += ksm(i,k),now%=mod;
65         cout<<now;
66         return 0;
67 
68     }
69     lg();
70 
71 }

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

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