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 + 10;
const long long mod = 1e9 + 7;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
vector<long long> g[N];
long long w[N], v[N];
long long ans = 0;
void solve(long long l, long long r, long long k, vector<long long> tmp) {
  if (l == r) {
    for (long long i = 1; i <= w[l] && i <= k; i++)
      ans = max(ans, tmp[k - i] + g[l][i]);
    ans = max(tmp[k], ans);
    return;
  }
  auto dp = tmp;
  long long mid = (l + r) >> 1;
  for (long long i = mid + 1; i <= r; i++)
    for (long long j = k; j >= w[i]; j--)
      dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
  solve(l, mid, k, dp);
  dp = tmp;
  for (long long i = l; i <= mid; i++)
    for (long long j = k; j >= w[i]; j--)
      dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
  solve(mid + 1, r, k, dp);
}
int main() {
  long long n = read(), k = read();
  for (long long i = 1; i <= n; i++) {
    long long k = read();
    g[i].resize(k + 1);
    for (long long j = 1; j <= k; j++) g[i][j] = read();
    for (long long j = 1; j <= k; j++) g[i][j] += g[i][j - 1];
    w[i] = k;
    v[i] = g[i][k];
  }
  vector<long long> dp(k + 1, -1e18);
  dp[0] = 0;
  solve(1, n, k, dp);
  cout << ans << endl;
}
