Problem: You are given n arrays that can have different sizes. You also have a table with w columns and n rows. The i-th array is placed horizontally in the i-th row. You can slide each array within its row as long as it occupies several consecutive cells and lies completely inside the table.

You need to find the maximum sum of the integers in the j-th column for each j from 1 to w independently.

<image> Optimal placements for columns 1, 2 and 3 are shown on the pictures from left to right.

Note that you can exclude any array out of a column provided it remains in the window. In this case its value is considered to be zero.

Input

The first line contains two integers n (1 ≤ n ≤ 10^{6}) and w (1 ≤ w ≤ 10^{6}) — the number of arrays and the width of the table.

Each of the next n lines consists of an integer l_{i} (1 ≤ l_{i} ≤ w), the length of the i-th array, followed by l_{i} integers a_{i1}, a_{i2}, …, a_{il_i} (-10^{9} ≤ a_{ij} ≤ 10^{9}) — the elements of the array.

The total length of the arrays does no exceed 10^{6}.

Output

Print w integers, the i-th of them should be the maximum sum for column i.

Examples

Input


3 3
3 2 4 8
2 2 5
2 6 3


Output


10 15 16 


Input


2 2
2 7 8
1 -8


Output


7 8 

Note

Illustration for the first example is in the statement.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long MOD = 1000000007;
long double EPS = 1e-9;
long long binpow(long long b, long long p, long long mod) {
  long long ans = 1;
  b %= mod;
  for (; p; p >>= 1) {
    if (p & 1) ans = ans * b % mod;
    b = b * b % mod;
  }
  return ans;
}
void pre() {}
long long dp[1000100];
long long n, w;
void addcontri(vector<long long>& arr, long long x) {
  vector<pair<long long, pair<long long, long long>>> sweep;
  for (int i = 0; i < x; i++) {
    int lo = i + 1;
    int hi = w - (x - i - 1);
    if (lo <= hi) {
      sweep.emplace_back(lo, make_pair(1, arr[i]));
      sweep.emplace_back(hi + 1, make_pair(2, arr[i]));
    }
  }
  long long len = w - x;
  if (len) {
    sweep.emplace_back(1, make_pair(1, 0));
    sweep.emplace_back(len + 1, make_pair(2, 0));
  }
  sweep.emplace_back(x + 1, make_pair(1, 0));
  sweep.emplace_back(w + 1, make_pair(2, 0));
  sort((sweep).begin(), (sweep).end());
  multiset<long long> avail;
  for (int i = 0; i < sweep.size() - 1; i++) {
    int j = i;
    while (j < sweep.size() && sweep[j].first == sweep[i].first) {
      if (sweep[j].second.first == 1) {
        avail.insert(sweep[j].second.second);
      } else {
        avail.erase(avail.find(sweep[j].second.second));
      }
      j++;
    }
    if (j < sweep.size() && !avail.empty()) {
      int lo = sweep[i].first;
      int hi = sweep[j].first;
      auto mini = avail.end();
      mini--;
      long long val = *mini;
      dp[lo] += val;
      dp[hi] -= val;
    }
    i = j - 1;
  }
}
void solve() {
  cin >> n >> w;
  for (long long i = 0; i < (n); ++i) {
    long long x;
    cin >> x;
    vector<long long> arr(x);
    for (long long i = 0; i < (x); ++i) cin >> arr[i];
    addcontri(arr, x);
  }
  for (long long i = 1; i <= w; ++i) {
    dp[i] += dp[i - 1];
    cout << dp[i] << " ";
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  pre();
  long long _t = 1;
  for (long long i = 1; i <= _t; i++) {
    solve();
  }
}