#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int MOD = (int)1e9 + 7;
int add(int x, int y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
int sub(int x, int y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
int mult(int x, int y) {
	return ((ll)x * y) % MOD;
}

const int N = 102;
const int S = N * N;
int dp[N][S];
int ANS;
int n, k;
int prod[N];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d", &n, &k);

	prod[0] = 1;
	for (int i = 1; i <= n; i++)
		prod[i] = mult(prod[i - 1], k + 1);

	dp[n][0] = 1;
	for (int p = n; p > 0; p--)
		for (int x = 0; x <= k * (n - p); x++)
			for (int y = 0; y <= k; y++) {
				if (y > p) {
					dp[p - 1][x] = add(dp[p - 1][x], dp[p][x]);
					continue;
				}
				int z = x + (x + y) / p;
				dp[p - 1][z] = add(dp[p - 1][z], dp[p][x]);
			}
	ANS = k * (k + 1) / 2;
	ANS = mult(ANS, prod[n - 1]);
	ANS = mult(ANS, n);

	for (int p = 1; p <= n; p++)
		for (int x = 0; x <= k * (n - p); x++)
			for (int y = 0; y <= k; y++) {
				if (y > p) continue;
				int z = (x + y) / p;
				ANS = sub(ANS, mult(z, mult(dp[p][x], prod[p - 1])));
			}
	printf("%d\n", ANS);

	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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).

Examples

Input

2 2


Output

10


Input

20 17


Output

983853488

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int MOD = (int)1e9 + 7;
int add(int x, int y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
int sub(int x, int y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
int mult(int x, int y) {
	return ((ll)x * y) % MOD;
}

const int N = 102;
const int S = N * N;
int dp[N][S];
int ANS;
int n, k;
int prod[N];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d", &n, &k);

	prod[0] = 1;
	for (int i = 1; i <= n; i++)
		prod[i] = mult(prod[i - 1], k + 1);

	dp[n][0] = 1;
	for (int p = n; p > 0; p--)
		for (int x = 0; x <= k * (n - p); x++)
			for (int y = 0; y <= k; y++) {
				if (y > p) {
					dp[p - 1][x] = add(dp[p - 1][x], dp[p][x]);
					continue;
				}
				int z = x + (x + y) / p;
				dp[p - 1][z] = add(dp[p - 1][z], dp[p][x]);
			}
	ANS = k * (k + 1) / 2;
	ANS = mult(ANS, prod[n - 1]);
	ANS = mult(ANS, n);

	for (int p = 1; p <= n; p++)
		for (int x = 0; x <= k * (n - p); x++)
			for (int y = 0; y <= k; y++) {
				if (y > p) continue;
				int z = (x + y) / p;
				ANS = sub(ANS, mult(z, mult(dp[p][x], prod[p - 1])));
			}
	printf("%d\n", ANS);

	return 0;
}

```