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;
const int MN = 3005;
int N, K;
long long ans = 0, f[MN];
vector<long long> a[MN];
inline void chkmax(long long &x, long long y) { x = max(x, y); }
inline void upd(int v, long long w) {
  for (int j = K; j >= v; --j) chkmax(f[j], f[j - v] + w);
}
inline void solve(int l, int r) {
  if (l == r) {
    for (int i = 0; i < a[l].size() && i <= K; ++i)
      chkmax(ans, f[K - i] + a[l][i]);
    return;
  }
  int mid = l + r >> 1;
  vector<long long> tmp;
  for (int i = 0; i <= K; ++i) tmp.emplace_back(f[i]);
  for (int i = mid + 1; i <= r; ++i)
    upd(a[i].size() - 1, a[i][a[i].size() - 1]);
  solve(l, mid);
  for (int i = 0; i <= K; ++i) f[i] = tmp[i];
  for (int i = l; i <= mid; ++i) upd(a[i].size() - 1, a[i][a[i].size() - 1]);
  solve(mid + 1, r);
  for (int i = 0; i <= K; ++i) f[i] = tmp[i];
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) {
    int m;
    scanf("%d", &m);
    a[i].emplace_back(0LL);
    for (int j = 1; j <= m; ++j) {
      int x;
      scanf("%d", &x);
      a[i].emplace_back(a[i][j - 1] + (long long)x);
    }
  }
  solve(1, N);
  printf("%lld\n", ans);
  return 0;
}
