#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

const ll mod = 1000000007LL;
const int limit = 10000;

bool ok[101][limit + 10];
ll num[101][limit + 10];
ll dp[101][limit + 10];

int main(void) {
	int n, k;
	cin >> n >> k;

	num[n][0] = 1LL;

	rrep(i, n) {
		rep(j, limit + 1) {
			if (num[i + 1][j] == 0) continue;
			rep(a, k + 1) {
				if (i + 1 < a) {
					num[i][j] = (num[i][j] + num[i + 1][j]) % mod;
					ll cur = (dp[i + 1][j] + 1LL * num[i + 1][j] * (a + j)) % mod;
					dp[i][j] = (dp[i][j] + cur) % mod;
				} else {
					const int add = (a + j) / (i + 1);
					num[i][j + add] = (num[i][j + add] + num[i + 1][j]) % mod;
					ll cur = (dp[i + 1][j] + 1LL * num[i + 1][j] * ((a + j) % (i + 1))) % mod;
					dp[i][j + add] = (dp[i][j + add] + cur) % mod;
				}
			}
		}
	}

	ll ans = 0LL;
	rep(i, limit + 1) ans = (ans + dp[0][i]) % mod;
	cout << ans << endl;

	return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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 <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

const ll mod = 1000000007LL;
const int limit = 10000;

bool ok[101][limit + 10];
ll num[101][limit + 10];
ll dp[101][limit + 10];

int main(void) {
	int n, k;
	cin >> n >> k;

	num[n][0] = 1LL;

	rrep(i, n) {
		rep(j, limit + 1) {
			if (num[i + 1][j] == 0) continue;
			rep(a, k + 1) {
				if (i + 1 < a) {
					num[i][j] = (num[i][j] + num[i + 1][j]) % mod;
					ll cur = (dp[i + 1][j] + 1LL * num[i + 1][j] * (a + j)) % mod;
					dp[i][j] = (dp[i][j] + cur) % mod;
				} else {
					const int add = (a + j) / (i + 1);
					num[i][j + add] = (num[i][j + add] + num[i + 1][j]) % mod;
					ll cur = (dp[i + 1][j] + 1LL * num[i + 1][j] * ((a + j) % (i + 1))) % mod;
					dp[i][j + add] = (dp[i][j + add] + cur) % mod;
				}
			}
		}
	}

	ll ans = 0LL;
	rep(i, limit + 1) ans = (ans + dp[0][i]) % mod;
	cout << ans << endl;

	return 0;
}
```