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 = 3007;
int t[MAXN];
vector<int> a[MAXN];
long long sa[MAXN];
long long dp[MAXN], odp[32][MAXN];
long long ans = 0;
int d = 0;
int n, k;
void calc(int l, int r) {
  if (l == r) {
    long long s = 0;
    ans = max(ans, s + dp[k]);
    for (int i = 0; i < t[l] && i < k; i++) {
      s += a[l][i];
      ans = max(ans, s + dp[k - i - 1]);
    }
    return;
  }
  int m = (l + r) / 2;
  memcpy(odp[d], dp, sizeof(long long) * (k + 1));
  d++;
  for (int i = m + 1; i <= r; i++) {
    for (int z = k; z >= t[i]; z--) {
      dp[z] = max(dp[z], dp[z - t[i]] + sa[i]);
    }
  }
  calc(l, m);
  d--;
  memcpy(dp, odp[d], sizeof(long long) * (k + 1));
  d++;
  for (int i = l; i <= m; i++) {
    for (int z = k; z >= t[i]; z--) {
      dp[z] = max(dp[z], dp[z - t[i]] + sa[i]);
    }
  }
  calc(m + 1, r);
  d--;
  memcpy(dp, odp[d], sizeof(long long) * (k + 1));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  memset(dp, 0, sizeof(long long) * (k + 1));
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    sa[i] = 0;
    for (int j = 0; j < t[i]; j++) {
      int x;
      cin >> x;
      sa[i] += x;
      a[i].push_back(x);
    }
  }
  calc(0, n - 1);
  cout << ans << '\n';
}
