Problem: Niwango bought a piece of land that can be represented as a half-open interval [0, X).

Niwango will lay out N vinyl sheets on this land. The sheets are numbered 1,2, \ldots, N, and they are distinguishable. For Sheet i, he can choose an integer j such that 0 \leq j \leq X - L_i and cover [j, j + L_i) with this sheet.

Find the number of ways to cover the land with the sheets such that no point in [0, X) remains uncovered, modulo (10^9+7). We consider two ways to cover the land different if and only if there is an integer i (1 \leq i \leq N) such that the region covered by Sheet i is different.

Constraints

* 1 \leq N \leq 100
* 1 \leq L_i \leq X \leq 500
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N X
L_1 L_2 \ldots L_N


Output

Print the answer.

Examples

Input

3 3
1 1 2


Output

10


Input

18 477
324 31 27 227 9 21 41 29 50 34 2 362 92 11 13 17 183 119


Output

134796357

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define sd(x) scanf("%d", &(x))
#define pii pair<int, int>
#define F first
#define S second
#define all(c) ((c).begin()), ((c).end())
#define sz(x) ((int)(x).size())
#define ld long double

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<", "<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"{";
	for(int i = 0;i < (int)v.size(); i++){
		if(i)os<<", ";
		os<<v[i];
	}
	os<<"}";
	return os;
}

#ifdef LOCAL
#define cerr cout
#else
#endif

#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif

const int N = 5e2 + 10, mod = 1e9 + 7;
int fact[N], invfact[N];
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (((ll) x) * y) % mod;}
inline int powr(int a, ll b){
	int x = 1 % mod;
	while(b){
		if(b & 1) x = mul(x, a);
		a = mul(a, a);
		b >>= 1;
	}
	return x;
}
inline int inv(int a){ return powr(a, mod - 2);}
void pre(){
	fact[0] = invfact[0] = 1;
	for(int i = 1;i < N; i++) fact[i] = mul(i, fact[i - 1]);
	invfact[N - 1] = inv(fact[N - 1]);
	for(int i = N - 2; i >= 1; i--) invfact[i] = mul(invfact[i + 1], i + 1);
	assert(invfact[1] == 1);
}

inline int C(int n, int k){
	if(n < k || k < 0) return 0;
	return mul(fact[n], mul(invfact[k], invfact[n - k]));
}

int dp[N][N], dp2[N][N];
int powers[N][N];

int num[N];
int main(){
	pre();
	for(int i = 0; i < N; i++){
		powers[i][0] = 1;
		for(int j = 1; j < N; j++) powers[i][j] = mul(i, powers[i][j - 1]);
	}
	int n = 100, L = 500; 
	sd(n); sd(L);
	for(int i = 1; i <= n; i++){
		int r = rand() % L + 1; 
		sd(r); 
		num[r]++;
	}
	L++;
	dp[0][0] = 1;
	for(int l = L; l >= 1; l--){
		memcpy(dp2, dp, sizeof dp);
		memset(dp, 0, sizeof dp);
		for(int jx = 0; jx <= L; jx++){
			for(int x = 0; x <= L; x++){
				int ways = dp2[jx][x];
				if(!ways) continue;
				for(int now = 0; now * l + jx <= L; now++){
					int total_len = now * l + jx;
					int total_num = now + x;
					
					dp[total_len][total_num] = add(dp[total_len][total_num], mul(ways, mul(invfact[now], powers[total_len - total_num * l][ num[l]])));
				}
			}
		}
	}
	int ans = 0;
	for(int r = 1; r <= L; r++){
		int val = mul(fact[r], dp[L][r]);
		if(r & 1) ans = add(ans, val);
		else ans = sub(ans, val);
	}
	printf("%d\n", ans);
}