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)

#include <iostream>
using namespace std;
int N, K, way[109][5009], dp[109][5009], mod = 1000000007;
int main() {
	cin >> N >> K;
	way[0][0] = 1;
	int mx = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j <= mx; j++) {
			for (int k = 0; k <= K; k++) {
				if (k > N - i) {
					way[i + 1][j] = (way[i + 1][j] + way[i][j]) % mod;
					dp[i + 1][j] = (dp[i + 1][j] + dp[i][j] + 1LL * (j + k) * way[i][j]) % mod;
				}
				else {
					int p = (j + k) / (N - i), q = (j + k) % (N - i);
					way[i + 1][j + p] = (way[i + 1][j + p] + way[i][j]) % mod;
					dp[i + 1][j + p] = (dp[i + 1][j + p] + dp[i][j] + 1LL * q * way[i][j]) % mod;
				}
			}
		}
		for (int j = 0; j <= 5000; j++) {
			if (way[i + 1][j]) mx = j;
		}
	}
	int ret = 0;
	for (int i = 0; i <= mx; i++) ret = (ret + dp[N][i]) % mod;
	cout << ret << endl;
	return 0;
}