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;

int main() {
  long n,q;
  cin >> n >> q;
  vector<long> s(n);
  vector<long> t(n);
  vector<long> x(n);
  vector<long> d(q);
  vector<vector<long>> aaa;
  map<long,long> bbb;
  for(long i=0;i<n;i++) cin >> s[i] >> t[i] >> x[i];
  for(long i=0;i<q;i++) cin >> d[i];

  for(long i=0;i<n;i++) {
    aaa.push_back({s[i]-x[i],0,x[i]});
    aaa.push_back({t[i]-x[i]-1,2,x[i]});
  }
  for(long i=0;i<q;i++) aaa.push_back({d[i],1});
  
  sort(aaa.begin(),aaa.end());
  for(long i=0;i<aaa.size();i++) {
    if(aaa[i][1]==0) {
      bbb[aaa[i][2]]++;
    } else if(aaa[i][1]==1) {
      if(bbb.size()==0) cout << -1 << endl;
      else cout << (*bbb.begin()).first << endl;
    } else {
      bbb[aaa[i][2]]--;
      if((*bbb.find(aaa[i][2])).second==0)  bbb.erase(bbb.find(aaa[i][2]));
    }
  }
}