You are given n non-decreasing arrays of non-negative numbers. 

Vasya repeats the following operation k times: 

  * Selects a non-empty array. 
  * Puts the first element of the selected array in his pocket. 
  * Removes the first element from the selected array. 



Vasya wants to maximize the sum of the elements in his pocket.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 3 000): the number of arrays and operations.

Each of the next n lines contain an array. The first integer in each line is t_i (1 ≤ t_i ≤ 10^6): the size of the i-th array. The following t_i integers a_{i, j} (0 ≤ a_{i, 1} ≤ … ≤ a_{i, t_i} ≤ 10^8) are the elements of the i-th array.

It is guaranteed that k ≤ ∑_{i=1}^n t_i ≤ 10^6.

Output

Print one integer: the maximum possible sum of all elements in Vasya's pocket after k operations.

Example

Input


3 3
2 5 10
3 1 2 3
2 1 20


Output


26
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3010;
int n, K;
long long ans, f[maxn];
vector<long long> V[maxn];
int main() {
  scanf("%d %d", &n, &K);
  for (int i = 1, k; i <= n; i++) {
    scanf("%d", &k);
    for (int j = 1, x; j <= k; j++) {
      scanf("%d", &x);
      if (j <= K) V[i].push_back((V[i].empty() ? 0 : V[i].back()) + x);
    }
  }
  memset(f, -0x3f, sizeof(f)), f[0] = 0;
  function<void(int, int)> solve = [&](int l, int r) {
    if (l == r) {
      for (int i = 0; i < min(K, (int)V[l].size()); i++) {
        ans = max(ans, f[K - i - 1] + V[l][i]);
      }
      return;
    }
    int mid = (l + r) >> 1;
    vector<long long> old;
    for (int i = 0; i < K; i++) {
      old.push_back(f[i]);
    }
    for (int i = mid + 1; i <= r; i++) {
      for (int j = K; j >= V[i].size(); j--) {
        f[j] = max(f[j], f[j - V[i].size()] + V[i].back());
      }
    }
    solve(l, mid);
    copy(old.begin(), old.end(), f);
    for (int i = l; i <= mid; i++) {
      for (int j = K; j >= V[i].size(); j--) {
        f[j] = max(f[j], f[j - V[i].size()] + V[i].back());
      }
    }
    solve(mid + 1, r);
  };
  solve(1, n), printf("%lld\n", ans);
  return 0;
}
