Problem: There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.

If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?

Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.

The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.

Output

Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo 109 + 7.

Examples

Input

3 2
2 4 5


Output

3


Input

4 3
7 8 9 10


Output

13


Input

4 0
5 10 20 21


Output

1

Note

In the first sample, we have three options: 

  * The first and second students form a group, and the third student forms a group. Total imbalance is 2 + 0 = 2. 
  * The first student forms a group, and the second and third students form a group. Total imbalance is 0 + 1 = 1. 
  * All three students form their own groups. Total imbalance is 0. 



In the third sample, the total imbalance must be 0, so each student must work individually.

Solution (in JAVA):

Solution:

import java.util.*;
import java.io.*;

public class GroupProjects2 {

	// https://codeforces.com/contest/626/problem/F
	
	// sol: https://codeforces.com/blog/entry/47764
	
	static long mod = (long)(1e9+7);
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("GroupProjects"));

		int t = 1;
		while (t-- > 0) {
			StringTokenizer st = new StringTokenizer(in.readLine());
			int n = Integer.parseInt(st.nextToken());
			int k = Integer.parseInt(st.nextToken());
			int[] arr = new int[n];
			st = new StringTokenizer(in.readLine());
			for (int i=0; i<n; i++) arr[i] = Integer.parseInt(st.nextToken());
			Arrays.parallelSort(arr);
			
			long[][] dp = new long[n][k+1];
				// index, number open, k so far
			
			// i = 0;
			dp[0][0] = 1;
			if (n > 1) dp[1][0] = 1;
			
			for (int i=1; i<n; i++) {
				long[][] temp = new long[n][k+1];
				for (int j=0; j<n; j++) {
					for (int x=0; x<=k; x++) {
						int newx = x + j * (arr[i] - arr[i-1]);
						if (newx > k) break;
						
						// create a new open group
						if (j+1 < n) {
							temp[j+1][newx] += dp[j][x];
							temp[j+1][newx] %= mod;
						}
						
						// create a new closed group
						temp[j][newx] += dp[j][x];
						temp[j][newx] %= mod;
						
						// add to an open group and close it
						if (j-1 >= 0) {
							temp[j-1][newx] += j * dp[j][x];
							temp[j-1][newx] %= mod;
						}

						// add to an open group but don't close it
						if (j > 0) {
							temp[j][newx] += j * dp[j][x];
							temp[j][newx] %= mod;
						}
						
//						System.out.println(i + " " + j + " " + newx + " " + temp[j][newx]);
					}
				}
				dp = temp;
			}
			
			long ans=0;
			for (int i=0; i<=k; i++) {
				ans += dp[0][i];
				ans %= mod;
			}
			
			System.out.println(ans);
		}
	}
}
	
/*
	public static long dp(int index, int open, int curk) {
		if (index >= n) {
			if (curk <= k && open == 0) return 1;
			return 0;
		}
		if (curk > k) return 0;
		if (memo[index][open][curk] != -1) return memo[index][open][curk];
		long ans=0;
		
		// create a new open group
		ans += dp(index+1, open+1, curk + (arr[index] - arr[index-1]) * open);
		
		// create a new closed group
		ans += dp(index+1, open, curk + (arr[index] - arr[index-1]) * open);
		
		// add to an open group and close it
		if (open-1 >= 0) ans += open*dp(index+1, open-1, curk + (arr[index] - arr[index-1]) * open);
		ans %= mod;
		
		// add to an open group but don't close
		if (open > 0) ans += open*dp(index+1, open, curk + (arr[index] - arr[index-1]) * open);
		ans %= mod;
		
		
		return memo[index][open][curk] = ans;
	}
*/