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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
const int N = 3000 + 1;
long long dp[N][N], su;
int n, k, temp, size1;
long long max(long long a, long long b) { return (a < b ? b : a); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> size1;
    su = 0;
    for (int j = 1; j <= k; ++j) dp[i][j] = dp[i - 1][j];
    for (int j = 1; j <= size1; ++j) {
      cin >> temp;
      su += temp;
      for (int itr = j; itr <= k; ++itr) {
        dp[i][itr] = max(dp[i][itr], su + dp[i - 1][itr - j]);
      }
    }
  }
  cout << dp[n][k];
  return 0;
}
