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;
int n, k;
vector<long long> dp;
vector<vector<long long> > a, sum;
long long ans;
void update(int l, int r) {
  for (int i = l; i <= r; ++i) {
    int sz = a[i].size();
    for (int j = k; j >= sz; --j) dp[j] = max(dp[j], dp[j - sz] + sum[i][sz]);
  }
}
void solve(int l, int r) {
  if (l == r) {
    int sz = a[l].size();
    for (int i = 0; i <= min(k, sz); ++i) ans = max(ans, sum[l][i] + dp[k - i]);
    return;
  }
  int mid = l + r >> 1;
  vector<long long> tmp(k + 1);
  for (int i = 0; i <= k; ++i) tmp[i] = dp[i];
  update(l, mid);
  solve(mid + 1, r);
  for (int i = 0; i <= k; ++i) dp[i] = tmp[i];
  update(mid + 1, r);
  solve(l, mid);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  dp.resize(k + 1);
  a.resize(n), sum.resize(n);
  for (int i = 0, m; i < n; ++i) {
    cin >> m;
    a[i].resize(m), sum[i].resize(m + 1);
    for (int j = 0; j < m; ++j) {
      cin >> a[i][j];
      sum[i][j + 1] = sum[i][j] + a[i][j];
    }
  }
  ans = 0;
  solve(0, n - 1);
  cout << ans << '\n';
  return 0;
}
