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;
int a[3007];
vector<long long> s[3007], tmp;
long long ans = 0LL;
void solve(int lx, int rx, vector<long long> g) {
  vector<long long> f;
  if (lx < rx) {
    int mid = (lx + rx) / 2;
    for (int i = 0; i <= k; i++) f.push_back(g[i]);
    for (int i = mid + 1; i <= rx; i++)
      for (int j = k; j >= a[i]; j--)
        f[j] = max(f[j], f[j - a[i]] + s[i][a[i]]);
    solve(lx, mid, f);
    f.clear();
    for (int i = 0; i <= k; i++) f.push_back(g[i]);
    for (int i = lx; i <= mid; i++)
      for (int j = k; j >= a[i]; j--)
        f[j] = max(f[j], f[j - a[i]] + s[i][a[i]]);
    solve(mid + 1, rx, f);
  } else {
    for (int i = 0; i <= a[lx] && i <= k; i++)
      ans = max(ans, s[lx][i] + g[k - i]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    long long sum = 0LL;
    s[i].push_back(sum);
    for (int j = 1; j <= a[i]; j++) {
      int x;
      scanf("%d", &x);
      sum += x;
      s[i].push_back(sum);
    }
  }
  for (int i = 0; i <= k; i++) tmp.push_back(0LL);
  solve(1, n, tmp);
  printf("%lld\n", ans);
  return 0;
}
