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
//解説に倣ってイベントソート
#include<bits/stdc++.h>
using namespace std;

void happen(int flag, int x, set<int> &s){
  if(flag==1) s.insert(x);
  else if(flag==-1) s.erase(x);
}

int main(){
  int n,q;
  cin>> n >> q;
  vector<vector<int>> event(2*n,vector<int>(3));
  vector<int> d(q);
  for(int i=0; i<n; i++){
    int s,t,x;
    cin>> s >> t >> x;
    event[i]={s-x,1,x};
    event[i+n]={t-x,-1,x};
  }
  for(int i=0; i<q; i++) cin>> d[i];
  
  sort(event.begin(),event.end());
  auto itr=event.begin();
  set<int> close;
  for(auto t:d){ //dはもともとソート済み
    while(itr!=event.end() && (*itr)[0]<=t){
      happen((*itr)[1],(*itr)[2],close);
      itr++;
    }
    if(close.empty()) cout<< -1 <<endl;
    else cout<< *close.begin() <<endl;
  }
}
//自力とかむり