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;

#define stx tuple<int, int, int>
#define xt pair<int, int>

int main() {
  int N, Q;
  cin >> N >> Q;
  priority_queue<stx, vector<stx>, greater<stx>> STX;
  int S, T, X, D[Q];
  for (int i = 0; i < N; i++) {
    cin >> S >> T >> X;
    STX.push(stx(S - X, T - X, X));
  }
  for (int i = 0; i < Q; i++) {
    cin >> D[i];
  }

  priority_queue<xt, vector<xt>, greater<xt>> XT;
  for (int i = 0; i < Q; i++) {
    while (!STX.empty() && get<0>(STX.top()) <= D[i]) {
      XT.push({get<2>(STX.top()), get<1>(STX.top())});
      STX.pop();
    }

    while (!XT.empty() && XT.top().second <= D[i]) XT.pop();

    if (XT.empty())
      cout << -1 << endl;
    else
      cout << XT.top().first << endl;
  }

  return 0;
}
