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>

#define debug(...) fprintf(stderr, __VA_ARGS__)

#ifndef AT_HOME
#define getchar() IO::myGetchar()
#define putchar(x) IO::myPutchar(x)
#endif

namespace IO {
	static const int IN_BUF = 1 << 23, OUT_BUF = 1 << 23;

	inline char myGetchar() {
		static char buf[IN_BUF], *ps = buf, *pt = buf;
		if (ps == pt) {
			ps = buf, pt = buf + fread(buf, 1, IN_BUF, stdin);
		}
		return ps == pt ? EOF : *ps++;
	}

	template<typename T>
	inline bool read(T &x) {
		bool op = 0;
		char ch = getchar();
		x = 0;
		for (; !isdigit(ch) && ch != EOF; ch = getchar()) {
			op ^= (ch == '-');
		}
		if (ch == EOF) {
			return false;
		}
		for (; isdigit(ch); ch = getchar()) {
			x = x * 10 + (ch ^ '0');
		}
		if (op) {
			x = -x;
		}
		return true;
	}

	inline int readStr(char *s) {
		int n = 0;
		char ch = getchar();
		for (; isspace(ch) && ch != EOF; ch = getchar())
			;
		for (; !isspace(ch) && ch != EOF; ch = getchar()) {
			s[n++] = ch;
		}
		s[n] = '\0';
		return n;
	}

	inline void myPutchar(char x) {
		static char pbuf[OUT_BUF], *pp = pbuf;
		struct _flusher {
			~_flusher() {
				fwrite(pbuf, 1, pp - pbuf, stdout);
			}
		};
		static _flusher outputFlusher;
		if (pp == pbuf + OUT_BUF) {
			fwrite(pbuf, 1, OUT_BUF, stdout);
			pp = pbuf;
		}
		*pp++ = x;
	}

	template<typename T>
	inline void print_(T x) {
		if (x == 0) {
			putchar('0');
			return;
		}
		static int num[40];
		if (x < 0) {
			putchar('-');
			x = -x;
		}
		for (*num = 0; x; x /= 10) {
			num[++*num] = x % 10;
		}
		while (*num){
			putchar(num[*num] ^ '0');
			--*num;
		}
	}

	template<typename T>
	inline void print(T x, char ch = '\n') {
		print_(x);
		putchar(ch);
	}

	inline void printStr_(const char *s, int n = -1) {
		if (n == -1) {
			n = strlen(s);
		}
		for (int i = 0; i < n; ++i) {
			putchar(s[i]);
		}
	}

	inline void printStr(const char *s, int n = -1, char ch = '\n') {
		printStr_(s, n);
		putchar(ch);
	}
}
using namespace IO;

const int N = 505, P = 1000000007;

int n, m, c[N], fac[N], inv[N], C[N][N], f[2][N][N];

int qpow(int a, int b) {
	int s = 1;
	for (; b; b >>= 1) {
		if (b & 1) {
			s = 1ll * s * a % P;
		}
		a = 1ll * a * a % P;
	}
	return s;
}

int main() {
	read(m), read(n);
	for (int i = 1; i <= m; ++i) {
		int x;
		read(x);
		++c[x];
	}
	fac[0] = 1;
	for (int i = 1; i <= n; ++i) {
		fac[i] = 1ll * fac[i - 1] * i % P;
	}
	inv[n] = qpow(fac[n], P - 2);
	for (int i = n; i; --i) {
		inv[i - 1] = 1ll * inv[i] * i % P;
	}
	for (int i = 0; i <= n; ++i) {
		C[i][0] = C[i][i] = 1;
		for (int j = 1; j < i; ++j) {
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
		}
	}
	int u = 0;
	f[u][0][0] = 1;
	for (int i = n; i; --i) {
		u ^= 1;
		for (int j = 0; j <= n; ++j) {
			for (int k = i * j; k <= n; ++k) {
				f[u][j][k] = 0;
				for (int l = j; ~l; --l) {
					f[u][j][k] = (f[u][j][k] + 1ll * f[u ^ 1][j - l][k - i * l] * inv[l]) % P;
				}
				f[u][j][k] = 1ll * f[u][j][k] * qpow(k - (i - 1) * j, c[i]) % P;
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= n; ++i) {
		for (int j = i; j <= n; ++j) {
			int t = 1ll * f[u][i][j] * fac[i] % P * C[n - j + 1][i] % P;
			if ((n - j) & 1) {
				ans = (ans + P - t) % P;
			} else {
				ans = (ans + t) % P;
			}
		}
	}
	print(ans);
}