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;
long long cur[3005], n, sum[3005];
vector<long long> a[3005];
long long second[3005], ans, k;
void go(long long l, long long r) {
  if (l == r) {
    long long pre = 0;
    for (int i = 0; i <= second[l]; i++) {
      pre += a[l][i];
      if (k >= i) ans = max(ans, cur[k - i] + pre);
    }
    return;
  }
  long long bef[k + 1];
  for (int i = 0; i <= k; i++) bef[i] = cur[i];
  long long mid = (l + r) / 2;
  for (int i = mid + 1; i <= r; i++)
    for (int j = k; j >= second[i]; j--)
      cur[j] = max(cur[j], cur[j - second[i]] + sum[i]);
  go(l, mid);
  for (int i = 0; i <= k; i++) cur[i] = bef[i];
  for (int i = l; i <= mid; i++)
    for (int j = k; j >= second[i]; j--)
      cur[j] = max(cur[j], cur[j - second[i]] + sum[i]);
  go(mid + 1, r);
  for (int i = 0; i <= k; i++) cur[i] = bef[i];
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> second[i];
    a[i].push_back(0);
    for (int j = 1; j <= second[i]; j++) {
      long long x;
      cin >> x;
      a[i].push_back(x);
      sum[i] += x;
    }
  }
  go(1, n);
  cout << ans << '\n';
  return 0;
}
