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<unordered_map>
#include<vector>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<set>
using namespace std;


int main(){
  int N, Q;
  cin >> N >> Q;
  
  vector<pair<int, pair<int, int>>> event;
  int s, t, x;
  for(int n = 0; n < N; n++){
    cin >> s >> t >> x;
    event.push_back(make_pair(s-x, make_pair(1, x)));
    event.push_back(make_pair(t-x, make_pair(-1, x)));
  }
  sort(event.begin(), event.end());
  
  int d;
  int arg = 0;
  
  set<int> X = {}; 
  
  for(int q=0; q<Q; q++){
    cin >> d;
    
    while (arg < 2*N && event[arg].first <= d){
      if(event[arg].second.first == 1){
        X.insert(event[arg].second.second);
      }else{
        X.erase(event[arg].second.second);
      }
      arg++;
    }
    
    if(X.empty()){
      cout << -1 << endl;
    } else{
      cout << *(X.begin()) << endl;
    }
  }

  return 0;
}