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 mxn = 3000;
int n, k;
long long a[mxn], first[mxn];
vector<long long> sol(int l, int r) {
  vector<long long> v;
  if (l == r) {
    cin >> first[l];
    v.resize(first[l] + 1);
    for (int i = 1; i <= first[l]; i++) {
      cin >> v[i];
      if (i <= k) a[l] += v[i];
      v[i] += v[i - 1];
    }
    v.resize(k + 1);
  } else {
    int mid = (l + r) / 2;
    vector<long long> vl = sol(l, mid), vr = sol(mid + 1, r);
    v.resize(k + 1);
    for (int i = l; i <= r; i++)
      for (int j = k; j >= first[i]; j--) {
        if (i <= mid)
          vr[j] = max(vr[j], a[i] + vr[j - first[i]]);
        else
          vl[j] = max(vl[j], a[i] + vl[j - first[i]]);
      }
    for (int i = 0; i <= k; i++) v[i] = max(vl[i], vr[i]);
  }
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  cout << sol(0, n - 1)[k] << '\n';
  return 0;
}
