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;
#define rep(i, m, n) for (int i = m; i < n; ++i)

int main() {
  int N, Q;
  cin >> N >> Q;
  using P = pair<int, int>;
  using PP = pair<P, int>;
  set<PP> st;
  set<int> run;
  rep(i, 0, N) {
    int S, T, X;
    cin >> S >> T >> X;
    st.insert(PP(P(S - X, X), 1));
    st.insert(PP(P(T - X, X), 0));
  }
  rep(i, 0, Q) {
    int D; cin >> D;
    while(!st.empty() && D >= (*st.begin()).first.first) {
      if((*st.begin()).second) run.insert((*st.begin()).first.second);
      else run.erase(run.find((*st.begin()).first.second));
      st.erase(st.begin());
      if(st.empty()) break;
    }
    if(run.empty()) cout << -1 << endl;
    else cout << *run.begin() << endl;
  }
  return 0;
}
