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 = 3005;
int n, k;
int t[MAXN];
int a[MAXN][MAXN];
long long dp[MAXN], cp[MAXN], s[MAXN];
long long sp[MAXN][MAXN];
long long ans;
void solve(int l, int r) {
  if (l == r) {
    for (int i = 0; i <= min(k, t[l]); ++i)
      ans = max(ans, sp[l][i] + dp[k - i]);
    return;
  }
  int mid = (l + r) / 2;
  vector<long long> dpc(k + 1);
  for (int i = 0; i <= k; ++i) dpc[i] = dp[i];
  for (int i = l; i <= mid; ++i)
    for (int j = k - t[i]; j >= 0; --j)
      dp[j + t[i]] = max(dp[j + t[i]], dp[j] + s[i]);
  solve(mid + 1, r);
  for (int i = 0; i <= k; ++i) dp[i] = dpc[i];
  dpc.clear();
  for (int i = mid + 1; i <= r; ++i)
    for (int j = k - t[i]; j >= 0; --j)
      dp[j + t[i]] = max(dp[j + t[i]], dp[j] + s[i]);
  solve(l, mid);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &t[i]);
    for (int j = 1; j <= t[i]; ++j) {
      int x;
      scanf("%d", &x);
      if (j > k) continue;
      sp[i][j] = sp[i][j - 1] + x;
      s[i] += x;
    }
  }
  solve(1, n);
  printf("%lld\n", ans);
  return 0;
}
