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;
long long read() {
  long long q = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    q = q * 10 + (ch - '0');
    ch = getchar();
  }
  return q * w;
}
void swap(long long &x, long long &y) {
  x = (x ^ y);
  y = (x ^ y);
  x = (x ^ y);
  return;
}
const long long N = 3000 + 95;
long long n, k, cnt, dp[N];
struct Node {
  long long tot, sum;
  vector<long long> v;
} a[N];
void Solve(long long l, long long r) {
  if (l == r) {
    for (long long i = 0; i < min(a[l].tot, k); i++)
      cnt = max(cnt, dp[k - i - 1] + a[l].v[i]);
    return;
  }
  long long mid = ((l + r) >> 1);
  long long f[N];
  memset(f, 0, sizeof(f));
  for (long long i = 0; i <= k; i++) f[i] = dp[i];
  for (long long i = (mid + 1); i <= r; i++)
    for (long long j = k; j >= a[i].tot; j--)
      dp[j] = max(dp[j], dp[j - a[i].tot] + a[i].sum);
  Solve(l, mid);
  for (long long i = 0; i <= k; i++) dp[i] = f[i];
  for (long long i = l; i <= mid; i++)
    for (long long j = k; j >= a[i].tot; j--)
      dp[j] = max(dp[j], dp[j - a[i].tot] + a[i].sum);
  Solve(mid + 1, r);
  for (long long i = 0; i <= k; i++) dp[i] = f[i];
  return;
}
int main() {
  n = read();
  k = read();
  for (long long i = 1; i <= n; i++) {
    a[i].tot = read();
    for (long long j = 1; j <= a[i].tot; j++) {
      long long x = read();
      a[i].sum += x;
      a[i].v.push_back(a[i].sum);
    }
  }
  Solve(1, n);
  cout << cnt << '\n';
  return 0;
}
