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;
inline long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
long long n, k;
vector<long long> a[3005];
long long sum[3005], len[3005];
vector<long long> solve(long long l, long long r) {
  if (l == r) {
    vector<long long> res(k + 1, 0);
    for (register long long i = (1); i <= (min(len[l], k)); ++i)
      res[i] = res[i - 1] + a[l][i];
    return res;
  }
  long long mid = l + r >> 1;
  vector<long long> fl = solve(l, mid), fr = solve(mid + 1, r), f(k + 1, 0);
  for (register long long i = (mid + 1); i <= (r); ++i)
    for (register long long j = (k); j >= (min(k, len[i])); --j)
      fl[j] = max(fl[j], fl[j - min(k, len[i])] + sum[i]);
  for (register long long i = (l); i <= (mid); ++i)
    for (register long long j = (k); j >= (min(k, len[i])); --j)
      fr[j] = max(fr[j], fr[j - min(k, len[i])] + sum[i]);
  for (register long long i = (0); i <= (k); ++i) f[i] = max(fl[i], fr[i]);
  return f;
}
signed main() {
  n = read(), k = read();
  for (register long long i = (1); i <= (n); ++i) {
    len[i] = read();
    a[i].resize(len[i] + 1);
    for (register long long j = (1); j <= (len[i]); ++j) {
      a[i][j] = read();
      if (j <= k) sum[i] += a[i][j];
    }
  }
  cout << solve(1, n)[k];
  return 0;
}
