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 ll = long long;

int main() {
  int N, Q;
  cin >> N >> Q;
  vector<tuple<ll, bool, ll>> events;
  for (int i = 0; i < N; ++i) {
    ll s, t, x;
    cin >> s >> t >> x;
    events.emplace_back(s - x, true, x);
    events.emplace_back(t - x, false, x);
  }
  sort(events.begin(), events.end());
  int j = 0;
  set<ll> xs;
  for (int i = 0; i < Q; ++i) {
    ll d;
    cin >> d;

    for (;; ++j) {
      if (j == events.size()) {
        cout << -1 << endl;
        break;
      }
      auto [t, to_add, x] = events[j];
      if (t > d) {
        if (xs.empty()) {
          cout << -1 << endl;
        } else {
          cout << *xs.begin() << endl;
        }
        break;
      }
      if (to_add) {
        xs.insert(x);
      } else {
        xs.erase(x);
      }
    }
  }
}