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 N = 3010;
vector<int> v[N];
long long dp[21][N], all[N], res;
int k;
void solve(int layer, int l, int r) {
  if (l == r) {
    res = max(res, dp[layer][k]);
    long long tmp = v[l][0];
    for (int i = 0; i < k && i < v[l].size(); tmp += v[l][++i])
      res = max(res, dp[layer][k - i - 1] + tmp);
    return;
  }
  int mid = (l + r) >> 1;
  auto push = [&](int l, int r) {
    for (int i = 1; i <= k; i++) dp[layer + 1][i] = dp[layer][i];
    for (int i = l; i <= r; i++)
      for (int j = k; j >= v[i].size(); j--)
        dp[layer + 1][j] =
            max(dp[layer + 1][j], dp[layer + 1][j - v[i].size()] + all[i]);
  };
  push(l, mid), solve(layer + 1, mid + 1, r), push(mid + 1, r),
      solve(layer + 1, l, mid);
}
int main() {
  int n;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    for (int j = 1, a; j <= t; j++)
      scanf("%d", &a), v[i].push_back(a), all[i] += a;
  }
  solve(0, 1, n), printf("%lld\n", res);
}
