// solution from usaco
#include<bits/stdc++.h>
#include<algorithm>
#include<random>
#include<chrono>
#include<cstdlib>
#include<ctime>
#include<numeric>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<iomanip>
#define int long long
#define ll long long
#define L LLONG_MAX
#define fi first
#define se second
#define pii pair<int,int>
#define sz(a) ((int)a.size())
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define REP(i,k,j) for(int i=k;i>=j;i--)
#define FORD(i,a) for(auto i:a)
#define rngdcl mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(l,r) ((l)+(rng()%(r-l+1)))
using namespace std;
const ll MOD = 1e9 + 7;
int n,l;
int a[102];
ll dp[102][102][1002][3];
void input()
{
cin >> n >> l;
for (int i = 1; i <= n; i++) cin >> a[i];
}
void solve()
{
if(n==1)
{
cout << 1;
return;
}
sort(a+1,a+n+1);
a[n+1]=10000;
dp[0][0][0][0]=1;
FOR(i,1,n) FOR(j,1,i) FOR(k,0,l) FOR(m,0,2)
{
int delta=(2*j-m)*(a[i+1]-a[i]);
if(delta>k || i+j+1-m>n) continue;
// Case 1
dp[i][j][k][m]+=dp[i-1][j-1][k-delta][m];
// Case 2
if(m) dp[i][j][k][m]+=(3-m)*dp[i-1][j-1][k-delta][m-1];
// Case 3
dp[i][j][k][m]+=(2*j-m)*dp[i-1][j][k-delta][m];
// Case 4
if(m==1) dp[i][j][k][m]+=2*j*dp[i-1][j][k-delta][m-1];
if(m==2)
{
if(i==n) dp[i][j][k][m]+=dp[i-1][j][k-delta][m-1];
else if(j>1) dp[i][j][k][m]+=(j-1)*dp[i-1][j][k-delta][m-1];
}
// Case 5
if(m==2)
{
if(i==n) dp[i][j][k][m]+=dp[i-1][j+1][k-delta][m];
else dp[i][j][k][m]+=j*(j-1)*dp[i-1][j+1][k-delta][m];
}
else if (m==1) dp[i][j][k][m]+=j*j*dp[i-1][j+1][k-delta][m];
else dp[i][j][k][m]+=j*(j+1)*dp[i-1][j+1][k-delta][m];
dp[i][j][k][m]%=MOD;
}
int ans=0;
FOR(i,0,l) ans+=dp[n][1][i][2];
cout << ans%MOD;
}
signed main()
{
//freopen(".inp", "r", stdin);
//freopen(".out", "w", stdout);
ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
input();
solve();
}
