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>
#include <algorithm>
using namespace std;
int main(){
	int N, Q; cin >> N >> Q; 
	pair<int, pair<int, int>> S[200010] = {}; 
	for(int i = 0; i < N; i++){
		cin >> S[i].second.first >> S[i].second.second >> S[i].first; 
	}
	sort(S, S+N); 
	int D[200010] = {}; 
	set<int> E; 
	for(int i = 0; i < Q; i++){
		cin >> D[i]; E.insert(D[i]);  
	}
	map<int, int> M; 
	for(int i = 0; i < N; i++){
		auto it1 = E.lower_bound(S[i].second.first - S[i].first); 
		auto it2 = E.lower_bound(S[i].second.second - S[i].first); 
		vector<int> used = {}; 
		for(auto it = it1; it != it2; it++){
			M[*it] = S[i].first; used.push_back(*it); 
		}
		for(int j = 0; j < used.size(); j++){
			E.erase(used[j]); 
		}
	}
	for(int i = 0; i < Q; i++){
		if(M[D[i]]) cout << M[D[i]] << endl; 
		else cout << -1 << endl; 
	}
}