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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int N = 3005;
long long dp[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  scanf("%d%d", &n, &k);
  for (int p = 1; p <= n; p++) {
    int x;
    scanf("%d", &x);
    long long s = 0;
    for (int j = 1; j <= k; j++) {
      dp[p][j] = dp[p - 1][j];
    }
    for (int i = 1; i <= x; i++) {
      int z;
      scanf("%d", &z);
      s += z;
      for (int j = i; j <= k; j++) {
        dp[p][j] = max(dp[p - 1][j - i] + s, dp[p][j]);
      }
    }
  }
  cout << dp[n][k] << endl;
  return 0;
}
