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:
// Don't place your source in a package


import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        //reading /writing file
        //Scanner sc=new Scanner(new File("input.txt"));
        //PrintWriter pr=new PrintWriter("output.txt");

        int T=1;
        for(int t=0;t<T;t++){
            int n=Int();int k=Int();
            int A[]=Arr(n);
            Solution sol=new Solution(out);
            sol.solution(A,k);
        }
        out.close();

    }

    public static int[] Arr(int n){
        int A[]=new int[n];
        for(int i=0;i<n;i++)A[i]=Int();
        return A;
    }
    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}








class Solution {
    PrintWriter out;
    int INF = Integer.MAX_VALUE;
    int MOD = 1000000007;

    public Solution(PrintWriter out) {
        this.out = out;
    }


    long dp[][][];
    int K;
    public void solution(int A[],int K) {
        this.K=K;
        Arrays.sort(A);
        int n =A.length;
        dp=new long[n/2+2][n+2][K+1];
        for(int i=0;i<dp.length;i++){
            for(int j=0;j<dp[0].length;j++){
                Arrays.fill(dp[i][j],-1);
            }
        }

        long res = dfs(A,0,0,0);
        out.println(res);
    }

    public long dfs(int A[],int left,int i,int k){
        if(k>this.K || left >= dp.length){
            return 0;
        }
        if(i>=A.length){
            if(left==0)return 1;
            else return 0;
        }
        if(dp[left][i][k]!=-1)return dp[left][i][k];

        long res =0;
        if(left==0){
            res+=dfs(A,1,i+1,k);//open
            res%=MOD;
            res+=dfs(A,0,i+1,k);//close on itself
            res%=MOD;
        }
        else{
            int cost = (A[i]-A[i-1])*left;
            //open
            res+=dfs(A,left+1,i+1,k+cost);
            res%=MOD;
            //close
            res+=left*dfs(A,left-1,i+1,k+cost);
            res%=MOD;
            //regular
            res+=(left + 1)*dfs(A,left,i+1,k+cost);
            res%=MOD;
        }
        dp[left][i][k]=res;
        return res;
    }
}