#include<bits/stdc++.h>
using namespace std;
// define
#define execute cerr << " Time: " << fixed << setprecision(6) << (1.0 * clock() / CLOCKS_PER_SEC) << "s\n";
#define ll long long
#define int long long
#define ld double
#define ii pair<int,int>
#define se second
#define fi first
#define iii pair<int,ii>
#define all(v) v.begin(),v.end()
#define bit(x,i) (((x)>>(i))&1LL)
#define flip(x,i) ((x)^(1LL<<(i)))
#define ms(d,x) memset(d,x,sizeof(d))
#define sitingfake 1
#define orz 1
//constant
const ll mod = 1e9 + 7;
const long long linf = 4557430888798830399;
const int inf = 1061109567;
const int maxarr = 1e6 + 5;
const int dx[] = {0,1,-1,0};
const int dy[] = {1,0,0,-1};
template<typename T> bool maximize(T &a, const T &b)
{
if(a < b) {a = b; return 1;}
return 0;
}
template<typename T> bool minimize(T &a, const T &b)
{
if(a > b) {a = b; return 1;}
return 0;
}
inline void Plus(ll & a ,ll b)
{
b %= mod;
a += b;
if(a >= mod) a -= mod;
if(a < 0) a += mod;
return;
}
inline void Mul(ll & a, ll b)
{
a %= mod; b %= mod;
a *= b;
a %= mod;
return;
}
//code
const int maxn = 200;
int a[maxn];
ll dp[2][105][1010][3];
int n , L;
void solve(void)
{
cin >> n >> L;
for(int i = 1; i <= n; i++)
{
cin >> a[i];
}
sort(a + 1, a + n + 1);
a[n + 1] = 10000;
dp[0][0][0][0] = 1;
for(int i = 1; i <= n ;i++)
{
int cur = i & 1;
int pre = ! cur;
ms(dp[cur] , 0);
for(int j = 1; j <= i ; j++)
{
for(int k = 0 ; k <= L ; k++)
{
for(int m = 0 ; m <= 2 ; m++)
{
int diff = (2 * j - m) * (a[i+1] - a[i]);
if(diff > k) continue;
Plus(dp[cur][j][k][m] , dp[pre][j-1][k - diff][m]);
if(m > 0)
{
Plus(dp[cur][j][k][m] , dp[pre][j-1][k - diff][m - 1] * (3 - m));
}
Plus(dp[cur][j][k][m] , dp[pre][j][k - diff][m] * (2 * j - m));
if(m == 1)
{
Plus(dp[cur][j][k][m] , dp[pre][j][k - diff][m - 1] * (2 * j));
}
if(m == 2)
{
if(i == n)
{
// when j = 1 note that all components merged into one when i != n we can not
// we cannot append it because when we create a new one and merge it later, it
//'s paradoxical
Plus(dp[cur][j][k][m] , dp[pre][j][k - diff][m - 1]);
}
else if(j > 1)
{
Plus(dp[cur][j][k][m] , dp[pre][j][k - diff][m - 1] * (j - 1));
}
}
if(m == 2)
{
if(i == n)// we can only merge two end of permutation when i = n && j = 1
{
Plus(dp[cur][j][k][m] , dp[pre][j + 1][k - diff][m]);
}
else
{
Plus(dp[cur][j][k][m] , dp[pre][j + 1][k - diff][m] * (j * (j - 1)));
}
}
else if(m == 1)
{
Plus(dp[cur][j][k][m] , dp[pre][j + 1][k - diff][m] * j * j);
}
else
{
Plus(dp[cur][j][k][m] , dp[pre][j + 1][k - diff][m] * j * (j + 1));
}
}
}
}
}
ll ans = 0;
for(int i = 0; i <= L ; i++)
{
Plus(ans , dp[n & 1][1][i][2]);
}
cout << ans;
}
/**
**/
signed main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#define task ""
if(fopen(task".inp","r"))
{
freopen(task".inp","r",stdin);
freopen(task".out","w",stdout);
}
int tc; tc = 1;
while(tc--) solve();
//execute;
}
