#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int mx = 1e6 + 5;
int n, w;
vector<int> vec[mx];
int l[mx];
set<pair<int, int>, greater<pair<int, int>>> st;
vector<int> tree[mx];
void init(int idx, int node, int l, int r) {
  if (l == r) {
    tree[idx][node] = vec[idx][l];
    return;
  }
  int mid = (l + r) / 2;
  init(idx, 2 * node, l, mid);
  init(idx, 2 * node + 1, mid + 1, r);
  tree[idx][node] = max(tree[idx][2 * node], tree[idx][2 * node + 1]);
}
int query(int idx, int node, int l, int r, int b, int e) {
  if (l > e || r < b) return -inf;
  if (l >= b && r <= e) return tree[idx][node];
  int mid = (l + r) / 2;
  int p1 = query(idx, 2 * node, l, mid, b, e);
  int p2 = query(idx, 2 * node + 1, mid + 1, r, b, e);
  return max(p1, p2);
}
int L[mx], R[mx];
long long tree2[mx * 4];
void update2(int node, int l, int r, int b, int x) {
  if (l == r) {
    tree2[node] = x;
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= b)
    update2(2 * node, l, mid, b, x);
  else
    update2(2 * node + 1, mid + 1, r, b, x);
  tree2[node] = tree2[2 * node] + tree2[2 * node + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> w;
  for (int i = 0; i < n; ++i) {
    L[i] = R[i] = 1;
    cin >> l[i];
    st.insert(make_pair(l[i], i));
    vec[i].resize(l[i] + 1);
    tree[i].resize((l[i] + 1) * 4);
    for (int j = 1; j <= l[i]; ++j) {
      cin >> vec[i][j];
    }
    init(i, 1, 1, l[i]);
  }
  for (int i = 1; i <= w; ++i) {
    long long ans = 0;
    for (set<pair<int, int>, greater<pair<int, int>>>::iterator itr =
             st.begin();
         itr != st.end(); ++itr) {
      if (min(i, w - i + 2) > (*itr).first) break;
      pair<int, int> p = *itr;
      L[p.second] = max(1, p.first - (w - i));
      R[p.second] = min(p.first, i);
      long long temp =
          query(p.second, 1, 1, l[p.second], L[p.second], R[p.second]);
      if (p.first < i || i + p.first <= w) temp = max(temp, 0LL);
      update2(1, 1, n, p.second + 1, temp);
    }
    ans = tree2[1];
    cout << ans << " ";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int mx = 1e6 + 5;
int n, w;
vector<int> vec[mx];
int l[mx];
set<pair<int, int>, greater<pair<int, int>>> st;
vector<int> tree[mx];
void init(int idx, int node, int l, int r) {
  if (l == r) {
    tree[idx][node] = vec[idx][l];
    return;
  }
  int mid = (l + r) / 2;
  init(idx, 2 * node, l, mid);
  init(idx, 2 * node + 1, mid + 1, r);
  tree[idx][node] = max(tree[idx][2 * node], tree[idx][2 * node + 1]);
}
int query(int idx, int node, int l, int r, int b, int e) {
  if (l > e || r < b) return -inf;
  if (l >= b && r <= e) return tree[idx][node];
  int mid = (l + r) / 2;
  int p1 = query(idx, 2 * node, l, mid, b, e);
  int p2 = query(idx, 2 * node + 1, mid + 1, r, b, e);
  return max(p1, p2);
}
int L[mx], R[mx];
long long tree2[mx * 4];
void update2(int node, int l, int r, int b, int x) {
  if (l == r) {
    tree2[node] = x;
    return;
  }
  int mid = (l + r) / 2;
  if (mid >= b)
    update2(2 * node, l, mid, b, x);
  else
    update2(2 * node + 1, mid + 1, r, b, x);
  tree2[node] = tree2[2 * node] + tree2[2 * node + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> w;
  for (int i = 0; i < n; ++i) {
    L[i] = R[i] = 1;
    cin >> l[i];
    st.insert(make_pair(l[i], i));
    vec[i].resize(l[i] + 1);
    tree[i].resize((l[i] + 1) * 4);
    for (int j = 1; j <= l[i]; ++j) {
      cin >> vec[i][j];
    }
    init(i, 1, 1, l[i]);
  }
  for (int i = 1; i <= w; ++i) {
    long long ans = 0;
    for (set<pair<int, int>, greater<pair<int, int>>>::iterator itr =
             st.begin();
         itr != st.end(); ++itr) {
      if (min(i, w - i + 2) > (*itr).first) break;
      pair<int, int> p = *itr;
      L[p.second] = max(1, p.first - (w - i));
      R[p.second] = min(p.first, i);
      long long temp =
          query(p.second, 1, 1, l[p.second], L[p.second], R[p.second]);
      if (p.first < i || i + p.first <= w) temp = max(temp, 0LL);
      update2(1, 1, n, p.second + 1, temp);
    }
    ans = tree2[1];
    cout << ans << " ";
  }
}

```