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;
inline long long read() {
  long long w = 0, r = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') r = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    w = (w << 1) + (w << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w * r;
}
int n, k;
int siz[N];
long long ans;
long long sum[N], ff[N];
vector<long long> a[N];
void solve(int l, int r, long long f[N]) {
  if (l == r) {
    long long val = 0;
    ans = max(ans, f[k]);
    for (int i = 1; i <= siz[l] && i <= k; i++) {
      val += a[l][i - 1];
      ans = max(ans, f[k - i] + val);
    }
    return;
  }
  int mid = (l + r) >> 1;
  long long dp[N];
  for (int i = 0; i <= k; i++) dp[i] = f[i];
  for (int i = l; i <= mid; i++) {
    for (int j = k; j >= siz[i]; j--) {
      dp[j] = max(dp[j], dp[j - siz[i]] + sum[i]);
    }
  }
  solve(mid + 1, r, dp);
  for (int i = 0; i <= k; i++) dp[i] = f[i];
  for (int i = mid + 1; i <= r; i++) {
    for (int j = k; j >= siz[i]; j--) {
      dp[j] = max(dp[j], dp[j - siz[i]] + sum[i]);
    }
  }
  solve(l, mid, dp);
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) {
    siz[i] = read();
    for (int j = 0; j < siz[i]; j++) {
      long long x = read();
      a[i].push_back(x);
      sum[i] += x;
    }
  }
  solve(1, n, ff);
  printf("%lld\n", ans);
  return 0;
}
