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 e = 3100;
int n, k;
vector<long long> v[e];
long long sum[e];
long long g[e], ans = 9999999999999 * (-1);
long long da(long long t1, long long t2) { return t1 > t2 ? t1 : t2; }
long long xi(long long t1, long long t2) { return t1 < t2 ? t1 : t2; }
void zl(int l, int r) {
  for (int i = l; i <= r; i++) {
    int wei = v[i].size();
    for (int j = k; j >= wei; j--) {
      g[j] = da(g[j - wei] + sum[i], g[j]);
    }
  }
}
void fz(int l, int r) {
  int mid = (l + r) / 2;
  long long w[e];
  if (l == r) {
    long long ss = 0;
    ans = da(ans, g[k]);
    for (int i = 0; i < k; i++) {
      if (i >= v[l].size()) break;
      ss += v[l][i];
      ans = da(ans, g[k - i - 1] + ss);
    }
  } else {
    for (int i = 0; i <= k; i++) w[i] = g[i];
    zl(mid + 1, r);
    fz(l, mid);
    for (int i = 0; i <= k; i++) g[i] = w[i];
    zl(l, mid);
    fz(mid + 1, r);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int q;
    long long a;
    cin >> q;
    for (int j = 1; j <= q; j++) {
      cin >> a;
      v[i].push_back(a);
      sum[i] += a;
    }
  }
  fz(1, n);
  cout << ans;
  return 0;
}
