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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

const int mod = 1000000007;

long long dp[101][4000];
int main() {
	int n, k;
	scanf("%d %d", &n, &k);
	long long ans = (n * (k * (k + 1) / 2)) % mod;
	for (int i = 0; i < n - 1; ++ i)
		(ans *= (k + 1)) %= mod;
	for (int i = 0; i < 4000; ++ i)
		dp[n][i] = 0;
	dp[n][0] = 1;
	for (int i = n; i > 0; -- i)
		for (int j = 0; j < 4000; ++ j)
			for (int a = 0; a <= k; ++ a)
				if (a <= i)
					(dp[i - 1][j + (a + j) / i] += dp[i][j]) %= mod;
				else
					(dp[i - 1][j] += dp[i][j]) %= mod;
	for (int j = 0; j < 4000; ++ j)
		ans = (ans + mod - j * dp[0][j] % mod) % mod;
	printf("%lld\n", ans);
	return 0;
}