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 maxn = 3123;
int n, k;
long long f[maxn], ans;
vector<int> a[maxn];
vector<long long> s[maxn];
void solve(int l, int r) {
  if (l == r) {
    for (int(i) = (0); (i) <= ((min(int(s[l].size()), k + 1)) - 1); (i)++)
      ans = max(ans, s[l][i] + f[i]);
    return;
  }
  vector<long long> pre(f, f + k + 1);
  int mi = l + r >> 1;
  for (int(i) = (l); (i) <= (mi); (i)++)
    for (int(j) = (int(s[i].size()) - 1); (j) <= (k); (j)++)
      f[j - s[i].size() + 1] = max(f[j - s[i].size() + 1], f[j] + s[i].back());
  solve(mi + 1, r);
  for (int(i) = (0); (i) <= ((k + 1) - 1); (i)++) f[i] = pre[i];
  for (int(i) = (mi + 1); (i) <= (r); (i)++)
    for (int(j) = (int(s[i].size()) - 1); (j) <= (k); (j)++)
      f[j - s[i].size() + 1] = max(f[j - s[i].size() + 1], f[j] + s[i].back());
  solve(l, mi);
  for (int(i) = (0); (i) <= ((k + 1) - 1); (i)++) f[i] = pre[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int(i) = (1); (i) <= (n); (i)++) {
    int t;
    cin >> t;
    a[i].resize(t + 1);
    s[i].resize(t + 1);
    for (int(j) = (1); (j) <= (t); (j)++) {
      cin >> a[i][j];
      s[i][j] = s[i][j - 1] + a[i][j];
    }
  }
  solve(1, n);
  cout << ans;
}
