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 = 3e3 + 5;
long long dp[20][maxn], c[maxn], a[maxn][maxn];
long long ans;
int n, k;
void init(int d) {
  for (int i = 0; i <= k; i++) dp[d][i] = dp[d - 1][i];
}
void upd(int d, int l, int r) {
  for (int i = l; i <= r; i++) {
    for (int j = k; j >= c[i]; j--) {
      dp[d][j] = max(dp[d][j], dp[d][j - c[i]] + a[i][c[i]]);
    }
  }
}
void sol(int d, int l, int r) {
  if (l == r) {
    ans = max(ans, dp[d][k]);
    for (int i = 1; i <= c[l] && i <= k; i++) {
      ans = max(ans, dp[d][k - i] + a[l][i]);
    }
    return;
  }
  int mid = l + r >> 1;
  init(d + 1);
  upd(d + 1, l, mid);
  sol(d + 1, mid + 1, r);
  init(d + 1);
  upd(d + 1, mid + 1, r);
  sol(d + 1, l, mid);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    for (int j = 1; j <= c[i] && j <= k; j++) {
      scanf("%lld", &a[i][j]);
      a[i][j] += a[i][j - 1];
    }
    for (int j = k + 1, d; j <= c[i]; j++) scanf("%d", &d);
  }
  sol(1, 1, n);
  printf("%lld\n", ans);
  return 0;
}
