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;
using LL = long long;
using ULL = unsigned long long;
int N, K;
vector<vector<LL>> A;
LL dp[3001][3001];
LL ans;
void proc(int l, int r) {
  if (l + 1 == r) {
    LL tmp = 0;
    for (int i = 0; i < (A[l].size()); i++)
      tmp = max(tmp, dp[1][K - i] + A[l][i]);
    ans = max(ans, tmp);
    return;
  }
  int m = (l + r) / 2;
  int z = r - l;
  for (int i = l; i < m; i++) {
    int k = A[i].size() - 1;
    LL a = A[i].back();
    for (int j = 0; j < k; j++) dp[z - 1][j] = dp[z][j];
    for (int j = k; j <= K; j++) dp[z - 1][j] = max(dp[z][j - k] + a, dp[z][j]);
    z--;
  }
  proc(m, r);
  z = r - l;
  for (int i = m; i < r; i++) {
    int k = A[i].size() - 1;
    LL a = A[i].back();
    for (int j = 0; j < k; j++) dp[z - 1][j] = dp[z][j];
    for (int j = k; j <= K; j++) dp[z - 1][j] = max(dp[z][j - k] + a, dp[z][j]);
    z--;
  }
  proc(l, m);
}
int main() {
  scanf("%d%d", &N, &K);
  A.resize(N);
  for (int i = 0; i < (N); i++) {
    int t;
    scanf("%d", &t);
    A[i].resize(t + 1);
    for (int j = 0; j < (t); j++) {
      int a;
      scanf("%d", &a);
      A[i][j + 1] = a;
    }
    A[i].resize(min((int)A[i].size(), K + 1));
    for (int j = 0; j < (A[i].size() - 1); j++) A[i][j + 1] += A[i][j];
  }
  ans = 0;
  for (int i = 0; i < (3001); i++) dp[N][i] = 0;
  proc(0, N);
  cout << ans << endl;
  return 0;
}
