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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
const int MAXN = 3010;
long long dp[2][MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int id = 1;
  for (int i = 1, ThxDem = n + 1; i < ThxDem; ++i) {
    int am;
    cin >> am;
    long long sum = 0;
    for (int j = 0, ThxDem = k + 1; j < ThxDem; ++j) dp[id][j] = dp[1 - id][j];
    for (int j = 1, ThxDem = am + 1; j < ThxDem; ++j) {
      int x;
      cin >> x;
      sum += x;
      for (int x = 0, ThxDem = k - j + 1; x < ThxDem; ++x)
        if (dp[1 - id][x] + sum > dp[id][x + j]) {
          dp[id][x + j] = dp[1 - id][x] + sum;
        }
    }
    id ^= 1;
  }
  cout << dp[1 - id][k] << "\n";
}
