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 N = 3e3 + 5;
long long f[20][N];
int a[N][N], m;
long long s[N];
long long ans = 0;
void solve(int dep, int l, int r) {
  if (l == r) {
    long long sum = 0;
    ans = max(ans, f[dep][m]);
    for (int i = 1; i <= min(m, a[l][0]); ++i)
      sum += a[l][i], ans = max(ans, sum + f[dep][m - i]);
    return;
  }
  int mid = l + r >> 1;
  for (int i = 0; i <= m; ++i) f[dep + 1][i] = f[dep][i];
  for (int i = mid + 1; i <= r; ++i)
    for (int j = m; j >= a[i][0]; --j)
      f[dep + 1][j] = max(f[dep + 1][j], f[dep + 1][j - a[i][0]] + s[i]);
  solve(dep + 1, l, mid);
  for (int i = 0; i <= m; ++i) f[dep + 1][i] = f[dep][i];
  for (int i = l; i <= mid; ++i)
    for (int j = m; j >= a[i][0]; --j)
      f[dep + 1][j] = max(f[dep + 1][j], f[dep + 1][j - a[i][0]] + s[i]);
  solve(dep + 1, mid + 1, r);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i][0];
    for (int j = 1; j <= a[i][0]; ++j) {
      int x;
      cin >> x;
      if (j <= m) a[i][j] = x, s[i] += x;
    }
  }
  solve(1, 1, n);
  cout << ans;
  return 0;
}
