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 ans, N, K, dp[3003];
vector<long long> pot[3003];
void divide(long long l, long long r) {
  if (l == r) {
    for (long long i = 0; i < pot[l].size() && i <= K; ++i)
      ans = max(ans, dp[K - i] + pot[l][i]);
    return;
  }
  long long mid = (l + r) / 2, tmp[3003];
  memcpy(tmp, dp, sizeof(tmp));
  for (long long i = l; i <= mid; ++i)
    for (long long j = K; j >= pot[i].size() - 1; --j)
      dp[j] = max(dp[j], dp[j - pot[i].size() + 1] + pot[i].back());
  divide(mid + 1, r);
  memcpy(dp, tmp, sizeof(dp));
  for (long long i = mid + 1; i <= r; ++i)
    for (long long j = K; j >= pot[i].size() - 1; --j)
      dp[j] = max(dp[j], dp[j - pot[i].size() + 1] + pot[i].back());
  divide(l, mid);
  memcpy(dp, tmp, sizeof(dp));
}
signed main() {
  ios::sync_with_stdio(0);
  memset(dp, -0x3f, sizeof(dp));
  cin >> N >> K;
  for (long long i = 1; i <= N; ++i) {
    pot[i].push_back(0);
    long long S;
    cin >> S;
    for (long long j = 1; j <= S; ++j) {
      long long x;
      cin >> x;
      pot[i].push_back(pot[i].back() + x);
    }
  }
  memset(dp, -0x3f, sizeof(dp));
  dp[0] = 0;
  ans = -1e18;
  divide(1, N);
  cout << ans;
  return 0;
}
