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<vector>
#include<algorithm>
#include<queue>
using namespace std;
int N,Q;
vector<pair<pair<int,int>,int> >E;
int main()
{
	cin>>N>>Q;
	for(int i=0;i<N;i++)
	{
		int S,T,X;cin>>S>>T>>X;
		S-=X,T-=X;
		E.push_back({{S,T},X});
	}
	sort(E.begin(),E.end());
	int id=0;
	priority_queue<pair<int,int> >P;
	for(int i=0;i<Q;i++)
	{
		int D;cin>>D;
		while(id<N&&E[id].first.first<=D)
		{
			P.push({-E[id].second,E[id].first.second});
			id++;
		}
		while(!P.empty()&&P.top().second<=D)P.pop();
		if(!P.empty())cout<<-P.top().first<<endl;
		else cout<<-1<<endl;
	}
}