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;

int main(){
  int N,Q;
  cin>>N>>Q;
  int I=1100100100;
  vector<pair<int,int>> p;
  for(int i=0;i<N;i++){
    int a,b,c;
    cin>>a>>b>>c;
    p.push_back(make_pair(a-c,c));
    p.push_back(make_pair(b-c,-1*c));
  }
  for(int i=0;i<Q;i++){
    int a;
    cin>>a;
    p.push_back(make_pair(a,I));
  }
  sort(p.begin(),p.end());
  priority_queue<int, vector<int>, greater<int>> pq;
  priority_queue<int, vector<int>, greater<int>> end;
  for(int i=0;i<2*N+Q;i++){
    int a,b;
    tie(a,b)=p[i];
    if(b<0){
      end.push(b*-1);
    }
    else if(b<I){
      pq.push(b);
    }
    else{
      while(!pq.empty()&&!end.empty()){
        if(pq.top()!=end.top()){
          break;
        }
        pq.pop();
        end.pop();
      }
      if(pq.empty()){
        cout<<"-1"<<endl;
      }
      else{
        cout<<pq.top()<<endl;
      }
    }
  }
}