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;
typedef pair<int,int>P;
priority_queue<P,vector<P>,greater<P> >p;
priority_queue<P,vector<P>,greater<P> >pq;

int main(){
  int n,q;
  int t[200009],x[200009];
  cin>>n>>q;
  for(int i=0;i<n;i++){
    int s;
    cin>>s>>t[i]>>x[i];
    t[i]-=x[i];
    p.push(P(s-x[i],i));
  }
  for(int i=0;i<q;i++){
    int d;
    cin>>d;
    while(!p.empty()){
      P now=p.top();
      int s=now.first;
      int idx=now.second;
      if(s>d)break;
      pq.push(P(x[idx],idx));
      p.pop();
    }
    while(!pq.empty()){
      P now=pq.top();
      int idx=now.second;
      if(t[idx]>d){
	cout<<x[idx]<<endl;
	break;
      }
      pq.pop();
    }
    if(pq.empty())cout<<-1<<endl;
  }
  return(0);
}
