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<int> A[3030];
long long S[3030];
long long ans;
void f(int l, int r, vector<long long> dp) {
  if (l + 1 == r) {
    long long s = 0;
    ans = max(ans, dp[K]);
    for (int i = 0; i < K && i < A[l].size(); i++) {
      s += A[l][i];
      ans = max(ans, dp[K - i - 1] + s);
    }
    return;
  }
  vector<long long> L = dp, R = dp;
  int mid = (l + r) / 2;
  for (int i = l; i < mid; i++) {
    int T = A[i].size();
    for (int k = K - T; k >= 0; k--) L[k + T] = max(L[k + T], L[k] + S[i]);
  }
  for (int i = mid; i < r; i++) {
    int T = A[i].size();
    for (int k = K - T; k >= 0; k--) R[k + T] = max(R[k + T], R[k] + S[i]);
  }
  f(l, mid, R);
  f(mid, r, L);
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; i++) {
    int T;
    scanf("%d", &T);
    A[i].reserve(T);
    for (int j = 0; j < T; j++) {
      int a;
      scanf("%d", &a);
      A[i].push_back(a);
      S[i] += a;
    }
  }
  vector<long long> dp(K + 1, 0);
  f(0, N, dp);
  printf("%lld\n", ans);
}
