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;
#define int long long

signed main() {
  int n,q;
  cin>>n>>q;
  int s,t,x;
  pair<int,pair<int,int>> k[n];
  for(int i=0;i<n;i++){
    cin>>s>>t>>x;
    k[i]=make_pair(s-x,make_pair(x,t-x));
  }  
  sort(k,k+n);
  priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> que;
  int d,p=0;
  for(int i=0;i<q;i++){
    cin>>d;
    while(p<n&&k[p].first<=d){
      que.push(k[p].second);
      p++;
    }
    while(!que.empty()&&que.top().second <=d)que.pop();
    if(que.empty())cout<<-1<<endl;
    else cout<<que.top().first<<endl;
  }
  


 


    return 0;
}
