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(){
	
	int N,Q;
	cin>>N>>Q;
	
	vector<pair<int,pair<int,int>>> V(N);
	for(int i=0;i<N;i++)scanf("%d %d %d",&V[i].second.first,&V[i].second.second,&V[i].first);
	sort(V.begin(),V.end());
	
	for(int i=0;i<N;i++){
		V[i].second.first-=V[i].first;
		V[i].second.second-=V[i].first;
	}
	
	vector<int> ans(Q,-1);
	
	set<int> S;
	vector<int> D(Q);
	for(int i=0;i<Q;i++){
		scanf("%d",&D[i]);
		S.insert(D[i]);
	}
	
	for(int i=0;i<N;i++){
		auto a = S.lower_bound(V[i].second.first);
		for(;a!=S.end();){
			if(*a<V[i].second.second){
				ans[lower_bound(D.begin(),D.end(),*a)-D.begin()]=V[i].first;
				a=S.erase(a);
			}
			else{
				break;
			}
		}
	}
	
	for(int i=0;i<Q;i++){
		printf("%d\n",ans[i]);
	}
	
	
	
	
	
	
	return 0;
}