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 read() {
  int X = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') X = X * 10 + c - '0', c = getchar();
  return X * w;
}
const int N = 3000 + 10;
int n, k;
vector<int> a[N];
long long s[N], dp[N], ans = 0;
void solve(int l, int r) {
  if (l == r) {
    int c = 0;
    long long s = 0;
    for (int i : a[l]) {
      ++c, s += i;
      ans = max(ans, dp[k - c] + s);
      if (k <= c) break;
    }
    return;
  }
  int mid = (l + r) >> 1;
  long long tmp[N];
  for (int i = 0; i <= k; ++i) tmp[i] = dp[i];
  for (int i = l; i <= mid; ++i)
    for (int j = k; j >= a[i].size(); --j)
      dp[j] = max(dp[j], dp[j - a[i].size()] + s[i]);
  solve(mid + 1, r);
  for (int i = 0; i <= k; ++i) dp[i] = tmp[i];
  for (int i = mid + 1; i <= r; ++i)
    for (int j = k; j >= a[i].size(); --j)
      dp[j] = max(dp[j], dp[j - a[i].size()] + s[i]);
  solve(l, mid);
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) {
    int t = read();
    while (t--) a[i].emplace_back(read()), s[i] += a[i].back();
  }
  solve(1, n);
  printf("%lld\n", ans);
  return 0;
}
