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>
inline int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
  return data * w;
}
const int N(3010);
std::vector<int> a[N];
int n, K, s[N];
long long f[N], sum[N], ans;
void Div(int l, int r) {
  if (l == r) {
    long long res = 0;
    for (int i = 0; i < (int)a[l].size() && i <= K; i++)
      ans = std::max(ans, res + f[K - i]), res += a[l][i];
    if ((int)a[l].size() <= K) ans = std::max(ans, res);
    return;
  }
  int mid = (l + r) >> 1;
  long long g[N];
  std::memcpy(g, f, (K + 1) << 3);
  for (int i = l; i <= mid; i++)
    for (int j = K; j >= s[i]; j--) f[j] = std::max(f[j], f[j - s[i]] + sum[i]);
  Div(mid + 1, r), std::memcpy(f, g, (K + 1) << 3);
  for (int i = mid + 1; i <= r; i++)
    for (int j = K; j >= s[i]; j--) f[j] = std::max(f[j], f[j - s[i]] + sum[i]);
  Div(l, mid);
}
int main() {
  n = read(), K = read();
  for (int i = 1; i <= n; i++) {
    a[i].reserve(s[i] = read());
    for (int j = 1, x; j <= s[i]; j++) a[i].push_back(x = read()), sum[i] += x;
  }
  Div(1, n), printf("%lld\n", ans);
  return 0;
}
