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 M,N;
    
    scanf("%d%d",&M,&N);
    
	vector<tuple<int,int,int>> vec;
	
	for(int i=0;i<M;i++){
	    
		int l,r,x;
		
		scanf("%d%d%d",&l,&r,&x);
		
		vec.emplace_back(x,l-x,r-x);
	}
	
	vector<int> Time(N);
	
	set<int> st;
	
	for(int i=0;i<N;i++){
	    
		scanf("%d",&Time[i]);
		
		st.insert(Time[i]);
	}
	
	sort(vec.begin(),vec.end());
	
	map<int,int> ans;
	
	for(int i=0;i<M;i++){
	    
		int cor=get<0>(vec[i]),l=get<1>(vec[i]),r=get<2>(vec[i]);
		
		auto it1=st.lower_bound(l);
		
		auto it2=st.lower_bound(r);
		
		vector<int> done;
		
		while(it1!=it2){
		    
			ans[*it1]=cor;
			
			done.push_back(*it1);
			
			it1++;
		}
		
		for(int x : done) st.erase(x);
	}
	for(int i=0;i<N;i++){
	    
		if(ans.count(Time[i])) cout<<ans[Time[i]]<<endl;
		
		else cout<<"-1"<<endl;
	}
	
	return 0;
	
}    
