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 argc, const char *argv[]) {
  int n, q;
  cin >> n >> q;
  vector<array<ll, 3>> vc(n);
  for (int i = 0; i < n; ++i) {
    ll s, t, x;
    cin >> s >> t >> x;
    vc[i][0] = x;
    vc[i][1] = s - x;
    vc[i][2] = t - x;
  }

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

  vector<ll> vd(q), vd2(q);
  for (int i = 0; i < q; ++i) {
    cin >> vd[i];
    vd2[i] = vd[i];
  }

  map<ll, ll> ans;
  for (int i = 0; i < n; ++i) {
    auto [x, s, t] = vc[i];
    auto it1 = lower_bound(vd.begin(), vd.end(), s);
    auto it2 = lower_bound(vd.begin(), vd.end(), t);
    for (auto it = it1; it != it2; it++) {
      ans[*it] = x;
    }

    vd.erase(it1, it2);
  }

  for (auto d : vd2) {
    if (ans.count(d) == 0) {
      cout << -1 << '\n';
    } else {
      cout << ans[d] << '\n';
    }
  }
}