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;

typedef pair<int, int> pii;
typedef long long ll;
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cerr << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U> void chmin(T& t, const U& u) { if (t > u) t = u; }
template<class T, class U> void chmax(T& t, const U& u) { if (t < u) t = u; }

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<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

template <unsigned int MOD>
struct ModInt {
	using uint = unsigned int;
	using ull = unsigned long long;
	using M = ModInt;

	uint v;

	ModInt(ll _v = 0) { set_norm(_v % MOD + MOD); }
	M& set_norm(uint _v) { //[0, MOD * 2)->[0, MOD)
		v = (_v < MOD) ? _v : _v - MOD;
		return *this;
	}

	explicit operator bool() const { return v != 0; }
	M operator+(const M& a) const { return M().set_norm(v + a.v); }
	M operator-(const M& a) const { return M().set_norm(v + MOD - a.v); }
	M operator*(const M& a) const { return M().set_norm(ull(v) * a.v % MOD); }
	M operator/(const M& a) const { return *this * a.inv(); }
	M& operator+=(const M& a) { return *this = *this + a; }
	M& operator-=(const M& a) { return *this = *this - a; }
	M& operator*=(const M& a) { return *this = *this * a; }
	M& operator/=(const M& a) { return *this = *this / a; }
	M operator-() const { return M() - *this; }
	M& operator++(int) { return *this = *this + 1; }
	M& operator--(int) { return *this = *this - 1; }

	M pow(ll n) const {
		if (n < 0) return inv().pow(-n);
		M x = *this, res = 1;
		while (n) {
			if (n & 1) res *= x;
			x *= x;
			n >>= 1;
		}
		return res;
	}

	M inv() const {
		ll a = v, b = MOD, p = 1, q = 0, t;
		while (b != 0) {
			t = a / b;
			swap(a -= t * b, b);
			swap(p -= t * q, q);
		}
		return M(p);
	}

	bool operator==(const M& a) const { return v == a.v; }
	bool operator!=(const M& a) const { return v != a.v; }
	friend ostream& operator<<(ostream& os, const M& a) { return os << a.v; }
};

using Mint = ModInt<1000000007>;

const int maxn = 110;
const int maxl = 510;
const int maxv = 1000010;

Mint fact[maxv], finv[maxv], inv[maxv];

void init() {
	fact[0] = 1;
	for (int i = 1; i < maxv; ++i) {
		fact[i] = fact[i-1] * i;
	}

	finv[maxv-1] = fact[maxv-1].inv();
	for (int i = maxv - 2; i >= 0; --i) {
		finv[i] = finv[i+1] * (i+1);
	}

	for (int i = 1; i < maxv; ++i) {
		inv[i] = finv[i] * fact[i-1];
	}
}

Mint dp[maxn][maxn][maxl];

int main() {
	init();

	int N, X;
	cin >> N >> X;
	V<int> L(N);
	rep(i, N) cin >> L[i];

	sort(ALL(L));
	reverse(ALL(L));

	dp[0][0][0] = 1;

	rep(i, N) {
		for (int j = 0; j * L[i] <= X && j <= i; ++j) {
			for (int k = 0; k < maxl; ++k) {
				if (!dp[i][j][k]) continue;
				//add interval
				if (k + L[i] <= X) {
					dp[i+1][j+1][k + L[i]] += dp[i][j][k] * Mint(j+1);
				}

				//included 
				if (j) {
					dp[i+1][j][k] += dp[i][j][k] * Mint(k - j * (L[i] - 1));
				}

				//elongate one interval

				if (j) {
					for (int w = 1; w < L[i]; ++w) {
						if (k + w <= X) {
							dp[i+1][j][k + w] += dp[i][j][k] * Mint(2 * j);
						}
					}
				}

				//merge two
				if (j >= 2) {
					for (int w = 0; w <= L[i] - 2; ++w) {
						if (k + w <= X) {
							dp[i+1][j-1][k + w] += dp[i][j][k] * Mint((L[i] - 1 - w) * (j - 1));
						}
					}
				}
			}
		}
	}

	Mint ans;
	for (int j = 1; j <= N; ++j) {
		ans += dp[N][j][X];
	}
	cout << ans << endl;

	return 0;
}