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 int long long

signed main() {
  int N, Q;
  cin >> N >> Q;
  vector<pair<pair<int, int>, int>> v;
  for (int i = 0; i < N; i++) {
    int s, t, x;
    cin >> s >> t >> x;
    v.push_back({{s - x, 1}, x});
    v.push_back({{t - x, 0}, x});
  }
  for (int i = 0; i < Q; i++) {
    int d;
    cin >> d;
    v.push_back({{d, 2}, 0});
  }

  sort(v.begin(), v.end());

  set<int> st;
  for (auto x : v) {
    int id = x.first.second;
    int pl = x.second;

    if (id == 1) {
      st.insert(pl);

    } else if (id == 0) {
      st.erase(pl);

    } else {
      if (st.size() == 0) {
        cout << -1 << endl;
      } else {
        cout << *st.begin() << endl;
      }
    }
  }

  return 0;
}
