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;
long long n, m, v[3001], w[3001], ans, dp[3001];
vector<long long> sum[3001];
inline long long read() {
  long long x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x;
}
void solve(long long l, long long r) {
  if (l == r) {
    for (register long long i = 0; i < (long long)sum[l].size(); ++i) {
      if (i > m) break;
      ans = max(ans, dp[m - i] + sum[l][i]);
    }
    return;
  }
  long long mid = (l + r) >> 1;
  long long tmp[3001];
  for (register long long i = 0; i <= m; ++i) tmp[i] = dp[i];
  for (register long long i = mid + 1; i <= r; ++i)
    for (register long long j = m; j >= v[i]; --j)
      dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
  solve(l, mid);
  for (register long long i = 0; i <= m; ++i) dp[i] = tmp[i];
  for (register long long i = l; i <= mid; ++i)
    for (register long long j = m; j >= v[i]; --j)
      dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
  solve(mid + 1, r);
}
signed main() {
  n = read(), m = read();
  for (register long long i = 1; i <= n; ++i) {
    long long p = read();
    sum[i].push_back(0);
    for (register long long j = 1; j <= p; ++j)
      sum[i].push_back(sum[i].back() + read());
    v[i] = p;
    w[i] = sum[i].back();
  }
  solve(1, n);
  printf("%lld\n", ans);
  return 0;
}
