Problem:
Score : 1000 points

Problem Statement

Consider the following game:

The game is played using a row of N squares and many stones.

First, a_i stones are put in Square i\ (1 \leq i \leq N).

A player can perform the following operation as many time as desired: "Select an integer i such that Square i contains exactly i stones. Remove all the stones from Square i, and add one stone to each of the i-1 squares from Square 1 to Square i-1."

The final score of the player is the total number of the stones remaining in the squares.

For a sequence a of length N, let f(a) be the minimum score that can be obtained when the game is played on a.

Find the sum of f(a) over all sequences a of length N where each element is between 0 and K (inclusive).
Since it can be extremely large, find the answer modulo 1000000007 (= 10^9+7).

Constraints

1 \leq N \leq 100

1 \leq K \leq N

Input

Input is given from Standard Input in the following format:

N K

Output

Print the sum of f(a) modulo 1000000007 (= 10^9+7).

Sample Input 1

2 2

Sample Output 1

10

There are nine sequences of length 2 where each element is between 0 and 2. For each of them, the value of f(a) and how to achieve it is as follows:

f(\{0,0\}): 0 (Nothing can be done)

f(\{0,1\}): 1 (Nothing can be done)

f(\{0,2\}): 0 (Select Square 2, then Square 1)

f(\{1,0\}): 0 (Select Square 1)

f(\{1,1\}): 1 (Select Square 1)

f(\{1,2\}): 0 (Select Square 1, Square 2, then Square 1)

f(\{2,0\}): 2 (Nothing can be done)

f(\{2,1\}): 3 (Nothing can be done)

f(\{2,2\}): 3 (Select Square 2)

Sample Input 2

20 17

Sample Output 2

983853488

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// By ksun48

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
 
LL dp[110][11000];
 
#define MOD 1000000007
LL powmod(LL a, LL n){
    if(n == 0) return 1;
    if(n % 2) return (a*powmod(a,n-1)) % MOD;
    LL c = powmod(a, n/2);
    return (c*c) % MOD;
}
LL inv(LL a){
    return powmod(a, MOD-2);
}
 
 
int main(){
    LL n, k;
    cin >> n >> k;
    for(LL N = 0; N <= n; N++){
        for(LL a = 0; a <= k*(n-N); a++){
            dp[N][a] = 0;
            if(N == 0) continue;
            for(LL val = 0; val <= k; val++){
                LL numadd = 0;
                LL newval = val + a;
                if(val <= N){
                    numadd = (val + a) / N;
                    newval = (val + a) % N;
                }
                dp[N][a] += dp[N-1][a + numadd];
                dp[N][a] %= MOD;
                dp[N][a] += newval * powmod(k+1, N-1);
                dp[N][a] %= MOD;
            }
        }
    }
    cout << dp[n][0] << endl;
}