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 nax = 3005;
long long dp[nax], sum[nax], a[nax][nax], answer = 0;
int sz[nax], n, k;
void rec(int l, int r) {
  if (l == r) {
    long long total = 0;
    for (int i = 0; i < int(sz[l] + 1); i++) {
      answer = max(answer, dp[k - i] + total);
      total += a[l][i];
    }
  } else {
    long long before_dp[nax];
    memcpy(before_dp, dp, sizeof dp);
    int m = l + (r - l) / 2;
    for (int row = l; row <= m; row++) {
      for (int ops = k; ops >= sz[row]; ops--) {
        dp[ops] = max(dp[ops], dp[ops - sz[row]] + sum[row]);
      }
    }
    rec(m + 1, r);
    memcpy(dp, before_dp, sizeof dp);
    for (int row = m + 1; row <= r; row++) {
      for (int ops = k; ops >= sz[row]; ops--) {
        dp[ops] = max(dp[ops], dp[ops - sz[row]] + sum[row]);
      }
    }
    rec(l, m);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < int(n); i++) {
    cin >> sz[i];
    for (int j = 0; j < int(sz[i]); j++) {
      int x;
      cin >> x;
      if (j < k) {
        a[i][j] = x;
      }
    }
    sz[i] = min(sz[i], k);
    for (int j = 0; j < int(sz[i]); j++) {
      sum[i] += a[i][j];
    }
  }
  rec(0, n - 1);
  cout << answer << endl;
}
