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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <set>
#define REP(i, n) for (int i = 0; i < (n); i++)

using namespace std;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int N, Q;
  cin >> N >> Q;
  vector<int> S(N), T(N), X(N);
  REP(i, N) cin >> S[i] >> T[i] >> X[i], S[i] -= X[i], T[i] -= X[i];
  vector<int> p(N);
  REP(i, N) p[i] = i;
  sort(p.begin(), p.end(), [&](int i, int j) {
    return X[i] < X[j];
  });
  vector<int> D(Q);
  set<pair<int, int>> st;
  REP(i, Q) cin >> D[i], st.emplace(D[i], i);
  vector<int> ans(Q, -1);
  for (int i : p) {
    auto it = st.lower_bound(make_pair(S[i], 0));
    while (it != st.end() && it->first < T[i]) {
      ans[it->second] = X[i];
      it = st.erase(it);
    }
  }
  REP(i, Q) cout << ans[i] << '\n';
}
