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;

using T = pair<int, pair<int, int>>;

vector<T> e;

const int INF = 1e9 + 100;

int main()
{
  int N, Q;
  cin >> N >> Q;
  set<int> s;
  s.insert(INF);
  for (int i = 0; i < N; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    e.emplace_back(a - c, make_pair(1, c));
    e.emplace_back(b - c, make_pair(-1, c));
  }
  e.emplace_back(INF, make_pair(0, 0));
  sort(e.begin(), e.end());
  int pos = 0;
  for (int i = 0; i < Q; i++) {
    int d;
    cin >> d;
    while (e[pos].first <= d) {
      if (e[pos].second.first == 1) {
        s.insert(e[pos].second.second);
      } else {
        s.erase(e[pos].second.second);
      }
      pos++;
    }
    int r = *(s.begin()) == INF ? -1 : *(s.begin());
    cout << r << endl;
  }
}