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>
const int MAXN = 3010;
int n, k, len[MAXN];
std::vector<long long> s[MAXN], f;
long long ans = 0;
void dp(int l, int r) {
  if (l == r) {
    for (int i = 0; i <= std::min(len[l] - 1, k - 1); i++)
      ans = std::max(ans, f[k - i - 1] + s[l][i]);
    return;
  }
  std::vector<long long> t = f;
  int mid = (l + r) >> 1;
  for (int i = l; i <= mid; i++)
    for (int j = k; j >= len[i]; j--)
      f[j] = std::max(f[j], f[j - len[i]] + s[i][len[i] - 1]);
  dp(mid + 1, r);
  f = t;
  for (int i = mid + 1; i <= r; i++)
    for (int j = k; j >= len[i]; j--)
      f[j] = std::max(f[j], f[j - len[i]] + s[i][len[i] - 1]);
  dp(l, mid);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", len + i);
    s[i].resize(len[i] + 2);
    for (int j = 0, x; j < len[i]; j++) {
      scanf("%d", &x);
      if (j != 0)
        s[i][j] = s[i][j - 1] + x;
      else
        s[i][j] = x;
    }
  }
  f.resize(k + 2);
  dp(1, n);
  printf("%lld\n", ans);
  return 0;
}
