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;
vector<int> V[3030];
int m[3030];
long long S[3030];
long long ans = 0;
long long dp[3030];
void f(int s, int e, int k) {
  if (s + 1 == e) {
    long long t = 0;
    for (int i = 0; i < m[s] && i <= k; ++i) {
      ans = max(ans, t + dp[k - i]);
      t += V[s][i];
    }
    if (m[s] <= k) ans = max(ans, t + dp[k - m[s]]);
    return;
  }
  int mid = s + e >> 1;
  long long _dp[k + 1];
  for (int i = 0; i <= k; ++i) _dp[i] = dp[i];
  for (int i = mid; i < e; ++i) {
    for (int j = k; j >= m[i]; --j) dp[j] = max(dp[j], dp[j - m[i]] + S[i]);
  }
  f(s, mid, k);
  for (int i = 0; i <= k; ++i) dp[i] = _dp[i];
  for (int i = s; i < mid; ++i) {
    for (int j = k; j >= m[i]; --j) dp[j] = max(dp[j], dp[j - m[i]] + S[i]);
  }
  f(mid, e, k);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> m[i];
    for (int j = 0; j < m[i]; ++j) {
      int x;
      cin >> x;
      V[i].push_back(x);
      S[i] += x;
    }
  }
  f(0, n, k);
  cout << ans;
}
