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;
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
long long ans;
int b[3005];
long long a[3005][3005], now[3005];
void solve(int l, int r) {
  if (l == r) {
    for (int i = 0; i <= b[l]; i++) chkmax(ans, a[l][i] + now[k - i]);
    return;
  }
  int mid = (l + r) / 2;
  long long tmp[3005];
  for (int i = 0; i <= k; i++) tmp[i] = now[i];
  for (int i = l; i <= mid; i++)
    for (int j = k; j >= b[i]; j--) chkmax(now[j], now[j - b[i]] + a[i][b[i]]);
  solve(mid + 1, r);
  for (int i = 0; i <= k; i++) now[i] = tmp[i];
  for (int i = mid + 1; i <= r; i++)
    for (int j = k; j >= b[i]; j--) chkmax(now[j], now[j - b[i]] + a[i][b[i]]);
  solve(l, mid);
}
int main() {
  n = readint();
  k = readint();
  int x, y;
  for (int i = 1; i <= n; i++) {
    x = readint();
    while (x--) {
      y = readint();
      if (b[i] < k) a[i][++b[i]] = y;
    }
    for (int j = 1; j <= b[i]; j++) a[i][j] += a[i][j - 1];
  }
  solve(1, n);
  printf("%lld\n", ans);
  return 0;
}
