#include <bits/stdc++.h>
using namespace std;
#define int long long
int dp[101][101][1001][3], a[101];
const int MOD = 1e9 + 7;
/*
dp[i][j][k][l] là số cách :
i - xét tới i
j - số lượng tplt
k - cost
l - điểm kết thúc
*/
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int n, l; cin >> n >> l;
for(int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
if(n == 1) cout << 1;
else{
a[n] = 10000; //inf for simplicity
if(a[1] - a[0] <= l) dp[1][1][a[1] - a[0]][1] = 2; //fill a[0] at one of the endpoints, there are 2 endpoints to fill.
if(2*(a[1] - a[0]) <= l) dp[1][1][2*(a[1] - a[0])][0] = 1; //fill a[0] in the middle, positions doesn't matter.
for(int i = 1; i < n; i++){
int diff = a[i + 1] - a[i]; //this thing is "INF" if i = n - 1.
for(int j = 1; j <= i; j++){
for(int k = 0; k <= l; k++){
for(int z = 0; z < 3; z++){
if(!dp[i][j][k][z]) continue; //this value does not exist
//First, we try to fill one of the ends
if(z < 2 && k + diff*(2*j - z - 1) <= l) //there are 2*j - z - 1 positions that we're supposed to "upgrade" (-1 because one of the positions is merged with the endpoints after this move)
{
if(i == n - 1)
{
dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] = (dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] + dp[i][j][k][z]*(2-z)*j)%MOD; //we have j con. comp. to choose to merge with
}
else if(z == 0 || j > 1) //otherwise this coincides with i == n - 1
{
dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] = (dp[i + 1][j][k + diff*(2*j - z - 1)][z + 1] + dp[i][j][k][z]*(2-z)*(j-z))%MOD; //can only merge with the con comp. that are not connected to ends.
}
if(k + diff*(2*j - z + 1) <= l) //now we create a new cc.
{
dp[i + 1][j + 1][k + diff*(2*j - z + 1)][z + 1] = (dp[i + 1][j + 1][k + diff*(2*j - z + 1)][z + 1] + dp[i][j][k][z]*(2-z))%MOD; //we can choose one of the ends to create
}
}
//Next, we dont fill the ends.
//Part 1 : Create new cc
if(k + diff*(2*j - z + 2) <= l) //2 new positions to "upgrade"
{
dp[i + 1][j + 1][k + diff*(2*j - z + 2)][z] = (dp[i + 1][j + 1][k + diff*(2*j - z + 2)][z] + dp[i][j][k][z])%MOD; //nothing new happens
}
//Part 2 : Stick to one cc
if(k + diff*(2*j - z) <= l) //no new positions to "upgrade"
{
dp[i + 1][j][k + diff*(2*j - z)][z] = (dp[i + 1][j][k + diff*(2*j - z)][z] + dp[i][j][k][z]*(2*j - z))%MOD; //we can merge in 2*j - z possible positions
}
//Part 3 : Merge two ccs together
if((k + diff*(2*j - z - 2) <= l) && (j >= 2) && (i == n - 1 || j > 2 || z < 2))
{
if(z == 0)
{
dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*j*(j-1))%MOD; //there are jP2 possible merges
}
if(z == 1)
{
dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*(j-1)*(j-1))%MOD; //there are (j-1)P2+(j-1) merges
}
if(z == 2){
if(i == n - 1){
dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z])%MOD; //there's only 1 place it can go.
}
else
{
dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] = (dp[i + 1][j - 1][k + diff*(2*j - z - 2)][z] + dp[i][j][k][z]*(j-2)*(j-1))%MOD; //there're (j-2)P2 + 2(j-2) possiblilities
}
}
}
}
}
}
}
int ans = 0;
for(int i = 0; i <= l; i++) ans = (ans + dp[n][1][i][2])%MOD; //sum the dp values for all possible sums
cout << ans;
}
}
