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>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
template <typename T, typename... Args>
void read(T &first, Args &...args) {
  read(first);
  read(args...);
}
template <typename T>
void write(T arg) {
  T x = arg;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename T, typename... Ts>
void write(T arg, Ts... args) {
  write(arg);
  if (sizeof...(args) != 0) {
    putchar(' ');
    write(args...);
  }
}
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int N = 3005;
int n, k;
int L[N];
vector<long long> sum[N];
long long dp[N], ans = 0;
void solve(int l, int r) {
  if (l == r) {
    for (int i = 0; i <= min(L[l], k); i++)
      ans = max(ans, dp[k - i] + sum[l][i]);
    return;
  }
  int mid = (l + r) >> 1;
  long long cpy[N];
  for (int i = 0; i <= k; i++) cpy[i] = dp[i];
  for (int i = mid + 1; i <= r; i++) {
    for (int j = k; j >= L[i]; j--)
      dp[j] = max(dp[j], dp[j - L[i]] + sum[i][L[i]]);
  }
  solve(l, mid);
  for (int i = 0; i <= k; i++) dp[i] = cpy[i];
  for (int i = l; i <= mid; i++) {
    for (int j = k; j >= L[i]; j--)
      dp[j] = max(dp[j], dp[j - L[i]] + sum[i][L[i]]);
  }
  solve(mid + 1, r);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> L[i];
    sum[i].push_back(0);
    for (int j = 1; j <= L[i]; j++) {
      int x;
      cin >> x;
      long long p = sum[i][j - 1] + x;
      sum[i].push_back(p);
    }
  }
  solve(1, n);
  cout << ans << endl;
}
