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 = 1e6 + 100;
int t[3005], n, k;
vector<long long> v[3005];
long long s[3005], ans;
void solve(int l, int r, long long a[]) {
  if (l == r) {
    long long sum = 0;
    ans = max(ans, a[k]);
    for (int i = 0; i < t[l]; i++) {
      sum += v[l][i];
      ans = max(ans, a[k - i - 1] + sum);
    }
    return;
  }
  long long b[3005], c[3005];
  int mid = (l + r) >> 1;
  for (int i = 0; i <= k; i++) b[i] = c[i] = a[i];
  for (int i = l; i <= mid; i++)
    for (int j = k; j >= t[i]; j--) c[j] = max(c[j], c[j - t[i]] + s[i]);
  for (int i = mid + 1; i <= r; i++)
    for (int j = k; j >= t[i]; j--) b[j] = max(b[j], b[j - t[i]] + s[i]);
  solve(l, mid, b);
  solve(mid + 1, r, c);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t[i]);
    for (int j = 0; j < t[i]; j++) {
      long long x;
      scanf("%d", &x);
      if (j < k) {
        s[i] += x;
        v[i].push_back(x);
      }
    }
    t[i] = min(t[i], k);
  }
  long long a[3005];
  for (int i = 0; i <= k; i++) a[i] = 0;
  solve(1, n, a);
  printf("%lld\n", ans);
  return 0;
}
