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 = 3000;
vector<int> a[Maxn + 5];
int n, k;
long long sum[Maxn + 5];
void update(long long *f, int w, long long v) {
  for (int i = k; i >= w; i--) {
    f[i] = max(f[i], f[i - w] + v);
  }
}
long long f[15][Maxn + 5];
long long ans;
void solve(int dep, int left, int right) {
  if (left == right) {
    long long sum = 0;
    for (int i = 0; i < (int)a[left].size(); i++) {
      ans = max(ans, f[dep][k - i] + sum);
      sum += a[left][i];
    }
    ans = max(ans, f[dep][k - a[left].size()] + sum);
    return;
  }
  int mid = (left + right) >> 1;
  memcpy(f[dep + 1], f[dep], sizeof f[dep]);
  for (int i = mid + 1; i <= right; i++) {
    update(f[dep + 1], a[i].size(), sum[i]);
  }
  solve(dep + 1, left, mid);
  memcpy(f[dep + 1], f[dep], sizeof f[dep]);
  for (int i = left; i <= mid; i++) {
    update(f[dep + 1], a[i].size(), sum[i]);
  }
  solve(dep + 1, mid + 1, right);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int len;
    scanf("%d", &len);
    a[i].reserve(len);
    for (int j = 0; j < len; j++) {
      int x;
      scanf("%d", &x);
      a[i].push_back(x);
    }
    while ((int)a[i].size() > k) {
      a[i].pop_back();
    }
    for (int j = 0; j < (int)a[i].size(); j++) {
      sum[i] += a[i][j];
    }
  }
  solve(0, 1, n);
  printf("%lld\n", ans);
  return 0;
}
