// JOI16_skyscraper
#include <bits/stdc++.h>
using namespace std;
#define int ll
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1000 * 1000 * 1000 + 7;
const int INF = 1e9 + 100;
const ll LINF = 1e18 + 100;
#ifdef DEBUG
#define dbg(x) cout << #x << " = " << (x) << endl << flush;
#define dbgr(s, f) { cout << #s << ": "; for (auto _ = (s); _ != (f); _++) cout << *_ << ' '; cout << endl << flush; }
#else
#define dbg(x) ;
#define dbgr(s, f) ;
#endif
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pb push_back
#define fr first
#define sc second
#define all(x) (x).begin(), (x).end()
#define endl '\n'
const int MAXN = 110;
const int MAXL = 1010;
int n, L, a[MAXN];
int dp[2][MAXN][MAXL][3];
inline void relax(int i, int j, int k, int l, ll val)
{
if (k < 0 || k >= MAXL) return;
dp[i][j][k][l] = (dp[i][j][k][l] + val) % mod;
}
int32_t main(void)
{
fast_io;
cin >> n >> L;
if (n == 1) return cout << "1\n", 0;
FOR(i, 1, n + 1) cin >> a[i];
sort(a + 1, a + n + 1);
// base
FOR(k, 0, MAXL)
{
dp[1][1][k][0] = 1;
dp[1][1][k][1] = 1;
}
// dp
FOR(i, 1, n)
{
int p = i % 2, c = 1 - p;
int dif = a[i + 1] - a[i];
memset(dp[c], 0, sizeof(dp[c]));
FOR(j, 1, i + 1)
{
FOR(k, 0, MAXL)
{
relax(c, j + 1, k + 2 * j * dif, 0, (j + 1) * (ll)dp[p][j][k][0]);
relax(c, j, k + 2 * j * dif, 0, 2 * j * (ll)dp[p][j][k][0]);
relax(c, j - 1, k + 2 * j * dif, 0, (j - 1) * (ll)dp[p][j][k][0]);
relax(c, j + 1, k + (2 * j - 1) * dif, 1, j * (ll)dp[p][j][k][1]);
relax(c, j + 1, k + 2 * j * dif, 1, dp[p][j][k][0]);
relax(c, j, k + (2 * j - 1) * dif, 1, (2 * j - 1) * (ll)dp[p][j][k][1]);
relax(c, j, k + 2 * j * dif, 1, dp[p][j][k][0]);
relax(c, j - 1, k + (2 * j - 1) * dif, 1, (j - 1) * (ll)dp[p][j][k][1]);
relax(c, j + 1, k + (2 * j - 2) * dif, 2, (j - 1) * (ll)dp[p][j][k][2]);
relax(c, j + 1, k + (2 * j - 1) * dif, 2, 2ll * dp[p][j][k][1]);
relax(c, j, k + (2 * j - 2) * dif, 2, (2 * j - 2) * (ll)dp[p][j][k][2]);
relax(c, j, k + (2 * j - 1) * dif, 2, 2ll * dp[p][j][k][1]);
relax(c, j - 1, k + (2 * j - 2) * dif, 2, (j - 1) * (ll)dp[p][j][k][2]);
}
}
if (i + 1 < n)
{
FOR(k, 0, MAXL) dp[c][1][k][2] = 0;
}
}
int ans = dp[n % 2][1][L][2];
ans = (mod + ans % mod) % mod;
cout << ans << endl;
return 0;
}
