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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int read() {
  int s = 0;
  char ch =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (ch < '0' || ch > '9')
    ch =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
  while (ch >= '0' && ch <= '9')
    s = s * 10 + (ch ^ 48),
    ch =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return s;
}
long long ans[3003], b[1000003];
using namespace std;
int main() {
  int n = read(), k = read();
  for (; n--;) {
    int m = read();
    for (int i = 1; i <= m; ++i) b[i] = b[i - 1] + read();
    m = min(m, k);
    for (int i = k; i >= 1; --i)
      for (int j = 1; j <= min(i, m); ++j)
        (ans[i] < ans[i - j] + b[j]) && (ans[i] = ans[i - j] + b[j]);
  }
  printf("%lld\n", ans[k]);
  return 0;
}
