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;

using query = pair<int, int>;
using event = pair<int, query>;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N; cin >> N;
  int Q; cin >> Q;
  vector<event> E;
  for (int i = 0; i < N; i++) {
    int S; cin >> S;
    int T; cin >> T;
    int X; cin >> X;
    E.emplace_back(S - X, query(0, X));
    E.emplace_back(T - X, query(1, X));
  }
  for (int i = 0; i < Q; i++) {
    int D; cin >> D;
    E.emplace_back(D, query(2, i));
  }
  sort(E.begin(), E.end());

  vector<int> ans(Q, -1);
  multiset<int> st;
  for (auto e: E) {
    int t = e.first;
    int q = e.second.first;
    int p = e.second.second;
    if (q == 0) st.insert(p);
    else if (q == 1) st.erase(st.lower_bound(p));
    else if (!st.empty()) ans[p] = *(st.begin());
  }

  for (int i = 0; i < Q; i++) {
    std::cout << ans[i] << '\n';
  }
  return 0;
}