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 GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
long long n, k;
long long dp[3001];
long long new_dp[3001];
long long ans = 0;
inline void relax(long long& a, long long b) { a = max(a, b); }
int32_t main() {
  ios_base::sync_with_stdio(false);
  scanf("%lld%lld", &n, &k);
  for (long long i = 0; i < n; ++i) {
    long long sz, sum = 0;
    scanf("%lld", &sz);
    memcpy(new_dp, dp, (k + 1) * sizeof(long long));
    for (long long j = 0; j < sz; ++j) {
      long long a;
      scanf("%lld", &a);
      sum += a;
      for (long long l = 0; l + j + 1 <= k; l++) {
        relax(new_dp[l + j + 1], dp[l] + sum);
      }
    }
    memcpy(dp, new_dp, (k + 1) * sizeof(long long));
  }
  printf("%lld", dp[k]);
  return 0;
}
