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,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
const long long mod = 1e9 + 7, ninf = -1e18, inf = 1e9;
const long long N = 3000 + 10;
long long arrays[N][N];
long long dp[N], currdp[N], su, tempdp[N], ma;
long long sizes[N], n, k, temp;
void solve() {
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) {
    cin >> sizes[i];
    for (long long j = 0; j < min(k, sizes[i]); ++j) cin >> arrays[i][j];
    for (long long j = min(k, sizes[i]); j < sizes[i]; ++j) cin >> temp;
  }
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i < n; ++i) {
    su = 0;
    for (long long j = 1; j <= min(sizes[i], k); ++j) {
      su += arrays[i][j - 1];
      currdp[j] = su;
    }
    for (long long j = sizes[i] + 1; j <= k; ++j) currdp[j] = currdp[j - 1];
    for (long long j = 1; j <= k; ++j) {
      ma = dp[j];
      for (long long itr = 1; itr <= min(j, sizes[i]); ++itr) {
        ma = max(ma, currdp[itr] + dp[j - itr]);
      }
      tempdp[j] = ma;
    }
    for (long long j = 0; j <= k; ++j) dp[j] = tempdp[j];
  }
  cout << dp[k] << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}
