There is an infinitely long street that runs west to east, which we consider as a number line.

There are N roadworks scheduled on this street. The i-th roadwork blocks the point at coordinate X_i from time S_i - 0.5 to time T_i - 0.5.

Q people are standing at coordinate 0. The i-th person will start the coordinate 0 at time D_i, continue to walk with speed 1 in the positive direction and stop walking when reaching a blocked point.

Find the distance each of the Q people will walk.

Constraints

* All values in input are integers.
* 1 \leq N, Q \leq 2 \times 10^5
* 0 \leq S_i < T_i \leq 10^9
* 1 \leq X_i \leq 10^9
* 0 \leq D_1 < D_2 < ... < D_Q \leq 10^9
* If i \neq j and X_i = X_j, the intervals [S_i, T_i) and [S_j, T_j) do not overlap.

Input

Input is given from Standard Input in the following format:


N Q
S_1 T_1 X_1
:
S_N T_N X_N
D_1
:
D_Q


Output

Print Q lines. The i-th line should contain the distance the i-th person will walk or -1 if that person walks forever.

Example

Input

4 6
1 3 2
7 13 10
18 20 13
3 4 2
0
1
2
3
5
8


Output

2
2
10
-1
13
-1
#include <bits/stdc++.h>
using namespace std;

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int n, q, s, t, d, roadwork[200200];
  vector<pair<int, int>> events;
  cin >> n >> q;
  for (int i = 1; i <= n; i++)
  {
    cin >> s >> t >> roadwork[i];
    events.push_back({s - roadwork[i], i});
    events.push_back({t - roadwork[i], -i});
  }

  sort(events.begin(), events.end());
  set<pair<int, int>> active;
  int i = 0;
  while (q--)
  {
    cin >> d;
    while (i < events.size() && events[i].first <= d)
    {
      int id = events[i].second;
      if (id > 0) active.insert({roadwork[id], id});
      else active.erase({roadwork[-id], -id});
      i++;
    }
    if (active.empty()) cout << "-1\n";
    else cout << active.begin()->first << '\n';
  }
}
