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;
int n, k, len[N], a[N][N], tmp[100010];
long long sum[N], dp[N], ans;
void insert(int v, long long w) {
  for (int i = k - v; i >= 0; i--) dp[i + v] = max(dp[i + v], dp[i] + w);
}
void solve(int l, int r) {
  if (l == r) {
    static long long pre[N];
    pre[0] = dp[0];
    for (int i = 1; i <= k; i++) pre[i] = max(pre[i - 1], dp[i]);
    long long now = 0;
    for (int i = 0; i <= len[l]; i++) {
      if (i) now += a[l][i];
      ans = max(ans, now + pre[k - i]);
    }
    return;
  }
  long long tmp[N];
  memcpy(tmp, dp, sizeof(dp));
  int mid = (l + r) >> 1;
  for (int i = mid + 1; i <= r; i++) insert(len[i], sum[i]);
  solve(l, mid);
  memcpy(dp, tmp, sizeof(tmp));
  for (int i = l; i <= mid; i++) insert(len[i], sum[i]);
  solve(mid + 1, r);
  memcpy(dp, tmp, sizeof(tmp));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> len[i];
    for (int j = 1; j <= len[i]; j++) cin >> tmp[j];
    len[i] = min(len[i], k);
    for (int j = 1; j <= len[i]; j++) a[i][j] = tmp[j], sum[i] += a[i][j];
  }
  solve(1, n);
  cout << ans;
  return 0;
}
