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 <algorithm>
#include <vector>
#include <map>

using namespace std;

struct E{
	int s,x;
	bool operator<(const E& e)const{
		if(s!=e.s)return s<e.s;
		return x<e.x;
	}
};

vector<E> ss;
vector<E> ts;
map<int,int> ms;

int main() {
	// your code goes here
	int n,q;
	cin>>n>>q;
	for(int i=0;i<n;i++){
		int x;
		E e1;
		E e2;
		cin>>e1.s>>e2.s>>x;
		e1.s-=x;
		e2.s-=x;
		e2.x=e1.x=x;
		ss.push_back(e1);
		ts.push_back(e2);
	}
	sort(ss.begin(),ss.end());
	sort(ts.begin(),ts.end());
	int p1=0,p2=0,p3=0;
	int d;
	cin>>d;
	while(p3<q){
		while(p1<ss.size() && ss[p1].s<=d){
			ms[ss[p1].x]++;
			p1++;
		}
		while(p2<ts.size() && ts[p2].s<=d){
			ms[ts[p2].x]--;
			if(ms[ts[p2].x]<=0)ms.erase(ts[p2].x);
			p2++;
		}
		if(ms.size()==0){
				cout<<"-1\n";
		}else{
			map<int,int>::iterator it=ms.begin();
			cout<<(*it).first<<"\n";
		}
		cin>>d;
		p3++;
	}
	return 0;
}