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<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int N_MAX=200000;
typedef pair<int,int> P;
typedef pair<int,P> PP;
int main()
{
  int n,q,d,i,qq;
  P p;
  PP s[N_MAX];
  priority_queue<P,vector<P>,greater<P>> que;
  cin >> n >> q;
  for(i=0;i<n;i++){
    cin >> s[i].first >> s[i].second.first >> s[i].second.second;
    s[i].first-=s[i].second.second;
    s[i].second.first-=s[i].second.second;
  }
  sort(s,s+n);
  i=0;
  for(qq=0;qq<q;qq++){
    cin >> d;
    for(;i<n;i++)
      if(s[i].first<=d){
        que.emplace(s[i].second.second,s[i].second.first);
      }else{
        break;
      }
    while(!que.empty()){
      p=que.top();
      if(p.second>d){
        cout << p.first << endl;
        break;
      }else{
        que.pop();
      }
    }
    if(que.empty())
      cout << "-1" << endl;
  }
  return 0;
}