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 MAP=pair<long,long>;
using Data=pair<MAP,long>;

int main(){
  int i,j,N,Q,cnt=0,a=0,b;
  long s,t,x;
  scanf("%d",&N);
  scanf("%d",&Q);

  vector<Data> event(2*N);
  set<long> se;
  for(i=0;i<N;i++){
    scanf("%ld",&s);
    scanf("%ld",&t);
    scanf("%ld",&x);
    event[cnt++]=Data(MAP(s-x,x),2);
    event[cnt++]=Data(MAP(t-x,x),1);
  }
  sort(event.begin(),event.end());
  cnt=0;
  for(i=0;i<Q;i++){
    scanf("%ld",&x);
    while(1){
      if(event[a].first.first<=x){
        switch(event[a].second){
          case 1:
            se.erase(event[a].first.second);
            break;
          case 2:
            se.insert(event[a].first.second);
            break;
        }
        a++;
      }else{
        break;
      }
    }
    if(se.empty()){
      printf("-1\n");
    }else{
      t=*se.upper_bound(0);
      printf("%d\n",t);
    }
  }
  return 0;
}