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;
const long long N = 3e3 + 7, NN = 1e18 + 7;
long long n, m, k, dp[20][N], s[N], len[N], ans;
vector<long long> a[N];
void dfs(long long l, long long r, long long cen) {
  if (l == r) {
    for (long long i = 1; i <= len[l]; i++) {
      ans = max(ans, dp[cen - 1][k - i] + a[l][i]);
    }
    return;
  }
  long long mi = (l + r) / 2;
  for (long long i = 0; i <= k; i++) dp[cen][i] = dp[cen - 1][i];
  for (long long i = mi + 1; i <= r; i++) {
    for (long long ii = k - len[i]; ii >= 0; ii--) {
      dp[cen][ii + len[i]] = max(dp[cen][ii + len[i]], dp[cen][ii] + s[i]);
    }
  }
  dfs(l, mi, cen + 1);
  for (long long i = 0; i <= k; i++) dp[cen][i] = dp[cen - 1][i];
  for (long long i = l; i <= mi; i++) {
    for (long long ii = k - len[i]; ii >= 0; ii--) {
      dp[cen][ii + len[i]] = max(dp[cen][ii + len[i]], dp[cen][ii] + s[i]);
    }
  }
  dfs(mi + 1, r, cen + 1);
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (long long t = 1; t <= n; t++) {
    scanf("%lld", &len[t]);
    a[t].push_back(0);
    for (long long i = 1; i <= len[t]; i++) {
      long long yyy;
      scanf("%lld", &yyy);
      if (i > k) {
        continue;
      }
      a[t].push_back(yyy);
      a[t][i] += a[t][i - 1];
    }
    len[t] = min(len[t], k);
    s[t] = a[t][len[t]];
  }
  dfs(1, n, 1);
  printf("%lld\n", ans);
}
