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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 20;
modint fact[max_n], factinv[max_n];
void init() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

//i番目,j個区間ある、和がx
modint dp[102][502];
modint cop[102][502];
modint rcop[102][503];
void solve() {
	int n, x; cin >> n >> x;
	vector<int> l(n);
	rep(i, n) {
		cin >> l[i];
	}
	sort(all(l), greater<int>());
	dp[1][l[0]] = 1;
	rep1(i, n - 1) {
		rep1(j, n) {
			rep1(k, x) {
				if (dp[j][k] == (modint)0)continue;
				//assert(k >= l[i] * j);
				//new segment
				if (k + l[i] <= x) {
					cop[j + 1][k + l[i]] += dp[j][k] * modint(j + 1);
				}
				//merge with 1 segment
				{
					//in 1 segment
					cop[j][k] += dp[j][k] *(modint) ((k-(l[i]-1)*(j-1) - l[i] + 1));
					//assert(k - (l[i] - 1)*(j - 1) - l[i] + 1 >= 0);
					//add length
					modint c = j * 2; c *= dp[j][k];
					for (int z = 1; z < l[i]; z++) {
						if (z + k > x)break;
						cop[j][k + z] += c;
					}
					/*rcop[j][k+1] += dp[j][k] * (modint)j*(modint)2;
					if (k + l[i] <= x) {
						rcop[j][k + l[i]] -= dp[j][k] * (modint)j*(modint)2;
					}*/
				}
				//merge with 2segment
				if (j >= 2) {
					modint c = j - 1;
					c *= dp[j][k];
					for (int z = 0; z <= l[i] - 2; z++) {
						if (k + z > x)break;
						cop[j - 1][k + z] += c * (modint)(l[i] - z - 1);
					}
				}
			}
		}
		rep1(j, n)rep1(k, x) {
			dp[j][k] = cop[j][k];
			cop[j][k] = 0;
		}
	}
	modint ans = 0;
	rep1(i, n)ans += dp[i][x];
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(7);
	//init();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}