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;
template <typename T>
inline T gi() {
  T f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const long long INF = 0x3f3f3f3f, N = 3003, M = N << 1;
long long n, k, p[N];
vector<long long> a[N];
long long sum[N];
vector<long long> dp;
long long ans;
void solve(long long l, long long r) {
  if (l == r) {
    long long now = 0;
    ans = max(ans, dp[k]);
    for (long long i = 1; i <= p[l]; i += 1) {
      if (i > k) return;
      now += a[l][i - 1];
      ans = max(ans, dp[k - i] + now);
    }
    return;
  }
  long long mid = (l + r) >> 1;
  vector<long long> tmp;
  tmp = dp;
  for (long long i = l; i <= mid; i += 1)
    for (long long j = k; j >= p[i]; j -= 1)
      dp[j] = max(dp[j], dp[j - p[i]] + sum[i]);
  solve(mid + 1, r);
  dp = tmp;
  for (long long i = mid + 1; i <= r; i += 1)
    for (long long j = k; j >= p[i]; j -= 1)
      dp[j] = max(dp[j], dp[j - p[i]] + sum[i]);
  solve(l, mid);
  return;
}
signed main() {
  n = gi<long long>(), k = gi<long long>();
  for (long long i = 1; i <= n; i += 1) {
    p[i] = gi<long long>();
    for (long long j = 1; j <= p[i]; j += 1) {
      long long x = gi<long long>();
      a[i].push_back(x);
      sum[i] += x;
    }
  }
  dp.resize(k + 1);
  solve(1, n);
  printf("%lld\n", ans);
  return 0;
}
