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 int maxN = 3100;
pair<int, long long> total[maxN];
long long dp[maxN];
vector<long long> a[maxN];
vector<long long> psum[maxN];
long long res = 0;
int n, k;
void solve(int l, int r) {
  if (l == r) {
    for (int i = 0; i <= k; ++i) {
      if (i - 1 >= (int)psum[l].size()) {
        break;
      }
      long long add = (i == 0) ? 0 : psum[l][i - 1];
      long long add2 = dp[k - i];
      res = max(res, add + add2);
    }
    return;
  }
  vector<long long> cur(k + 1);
  for (int i = 0; i <= k; ++i) {
    cur[i] = dp[i];
  }
  int mid = (l + r) / 2;
  for (int i = l; i <= mid; ++i) {
    for (int j = k; j >= 0; --j) {
      if (j + total[i].first <= k && dp[j] != -1) {
        dp[j + total[i].first] =
            max(dp[j + total[i].first], dp[j] + total[i].second);
      }
    }
  }
  solve(mid + 1, r);
  for (int i = 0; i <= k; ++i) {
    dp[i] = cur[i];
  }
  for (int i = mid + 1; i <= r; ++i) {
    for (int j = k; j >= 0; --j) {
      if (j + total[i].first <= k && dp[j] != -1) {
        dp[j + total[i].first] =
            max(dp[j + total[i].first], dp[j] + total[i].second);
      }
    }
  }
  solve(l, mid);
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    int len;
    cin >> len;
    a[i].resize(len);
    psum[i].resize(len);
    long long sum = 0;
    for (int j = 0; j < len; ++j) {
      cin >> a[i][j];
      sum += a[i][j];
    }
    psum[i] = a[i];
    for (int j = 1; j < len; ++j) {
      psum[i][j] = psum[i][j - 1] + a[i][j];
    }
    total[i] = make_pair(len, sum);
  }
  memset(dp, -1, sizeof(dp));
  dp[0] = 0;
  solve(0, n - 1);
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  t = 1;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}
