#define _GLIBCXX_DEBUG #include #include using namespace std; using mint = atcoder::modint998244353; using T = tuple; mint memo[101][5001][101]; int main() { for(int i = 0; i <= 100; i++) for(int j = 0; j <= 5000; j++) for(int k = 0; k <= 100; k++) memo[i][j][k] = -1; auto f = [&](auto f, int n, int m, int k) { if(memo[n][m][k] != -1) return memo[n][m][k]; if(m < 0 or k < 0 or n * k < m) return memo[n][m][k] = 0; if(n == 0) return memo[n][m][k] = !m; return memo[n][m][k] = f(f, n - 1, m, k) + f(f, n, m - n, k - 1); }; int n, k; cin >> n >> k; vector a(n, 0), cnt(n, 0); for(auto& v : a) cin >> v; if(k > n * (n - 1) / 2) { puts("0"); return 0; } sort(a.begin(), a.end()); int now = 0; for(int i = 0; i < n - 1; i++) { cnt[now]++; if(a[i] != a[i + 1]) now++; } cnt[now]++; vector dp(n + 1, vector(k + 1, mint(0))); dp[0][0] = 1; int cum = 0; for(int i = 0; i < n; i++) { for(int j = 0; j <= k; j++) for(int l = 0; l <= min(cnt[i] * cum, j); l++) dp[i + 1][j] += dp[i][j - l] * f(f, cnt[i], l, cum); cum += cnt[i]; } cout << dp.back()[k].val() << '\n'; return 0; }