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;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;
const int N = 3005, LOG = 15;
VI a[N];
ll dp[LOG][N], sum[N];
void push(int l, int r, int d, const int k) {
  for (int j = 0; j <= k; j++) {
    dp[d][j] = dp[d - 1][j];
  }
  for (int i = l; i < r; i++) {
    int sz = a[i].size();
    for (int j = k; j >= sz; j--) {
      dp[d][j] = max(dp[d][j], dp[d][j - sz] + sum[i]);
    }
  }
}
ll ans = 0;
void solve(int l, int r, int d, const int k) {
  if (l + 1 == r) {
    d--;
    const auto &vec = a[l];
    ans = max(ans, dp[d][k]);
    ll cur = 0;
    for (int i = 1; i <= (int)vec.size() && i <= k; i++) {
      cur += vec[i - 1];
      ans = max(ans, dp[d][k - i] + cur);
    }
    return;
  }
  int m = (l + r) / 2;
  push(m, r, d, k);
  solve(l, m, d + 1, k);
  push(l, m, d, k);
  solve(m, r, d + 1, k);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int sz;
    cin >> sz;
    a[i].resize(sz);
    for (int &x : a[i]) {
      cin >> x;
      sum[i] += x;
    }
  }
  solve(0, n, 1, k);
  cout << ans << endl;
  return 0;
}
