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(2)
#pragma GCC optimize(3)
using namespace std;
const int N = 2e5 + 10;
int n, k, c[N];
long long a[3003][3003], ans;
long long dp[22][3003];
void gao(int l, int r, int d) {
  if (l == r) {
    for (int i = 0; i <= min(k, c[l]); i++) {
      ans = max(ans, dp[d][k - i] + a[l][i]);
    }
    return;
  }
  int mid = l + r >> 1;
  memcpy(dp[d + 1], dp[d], sizeof(dp[d]));
  for (int i = mid + 1; i <= r; i++) {
    if (c[i] <= k) {
      for (int j = k; j >= 0; j--) {
        if (j - c[i] >= 0)
          dp[d + 1][j] = max(dp[d + 1][j], dp[d + 1][j - c[i]] + a[i][c[i]]);
      }
    }
  }
  gao(l, mid, d + 1);
  memcpy(dp[d + 1], dp[d], sizeof(dp[d]));
  for (int i = l; i <= mid; i++) {
    if (c[i] <= k) {
      for (int j = k; j >= 0; j--) {
        if (j - c[i] >= 0)
          dp[d + 1][j] = max(dp[d + 1][j], dp[d + 1][j - c[i]] + a[i][c[i]]);
      }
    }
  }
  gao(mid + 1, r, d + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    for (int j = 1; j <= c[i]; j++) {
      int x;
      cin >> x;
      if (j <= k) a[i][j] = x + a[i][j - 1];
    }
  }
  gao(1, n, 0);
  cout << ans << '\n';
  return 0;
}
