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;
vector<long long> dp;
vector<vector<long long>> a;
vector<long long> sum;
vector<long long> sz;
void relax(long long& a, long long b) { a = max(a, b); }
long long ans = 0;
long long n, k;
void rec(long long l, long long r) {
  if (l == r - 1) {
    long long sum = 0;
    for (long long i = 1; i <= min(k, sz[l]); i++) {
      sum += a[l][i - 1];
      relax(ans, sum + dp[k - i]);
    }
    return;
  }
  vector<long long> dp_copy = dp;
  long long mid = (l + r) / 2;
  for (long long i = l; i < mid; i++) {
    for (long long j = k - sz[i]; j >= 0; j--) {
      relax(dp[j + sz[i]], dp[j] + sum[i]);
    }
  }
  rec(mid, r);
  dp = dp_copy;
  for (long long i = mid; i < r; i++) {
    for (long long j = k - sz[i]; j >= 0; j--) {
      relax(dp[j + sz[i]], dp[j] + sum[i]);
    }
  }
  rec(l, mid);
  dp = dp_copy;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  a.resize(n);
  sum.resize(n);
  sz.resize(n);
  dp.resize(k + 1);
  for (long long i = 0; i < n; i++) {
    cin >> sz[i];
    a[i].resize(sz[i]);
    for (long long j = 0; j < sz[i]; j++) {
      cin >> a[i][j];
      sum[i] += a[i][j];
    }
  }
  rec(0, n);
  cout << ans << endl;
  return 0;
}
