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;

typedef long long lint;
typedef pair<int, int> pii;
typedef pair<int, pii> pip;

const int INF = 2e9;
const int MAX = 2e5+10;
const lint LNF = 2e18;

int n, q; // s-x가 작은것 부터

int main(){
	// ios::sync_with_stdio(0); cin.tie(0);

	map<int, vector<int>> inc;
	cin>>n>>q;
	for(int i=1; i<=n; i++){
		int s,t,x; cin>>s>>t>>x;
		inc[s-x].push_back(x), inc[t-x].push_back(-x);
	}

	// for(auto p:inc){
	// 	cout<<p.first<<": ";
	// 	for(int x:p.second) cout<<x<<' ';
	// 	cout<<'\n';
	// }
	// cout<<'\n';
	
	for(int i=1; i<=q; i++){
		static multiset<int> S;
		int d; cin>>d;
		while(!inc.empty() && inc.begin()->first<=d){
			vector<int> &V = inc.begin()->second;
			for(int x:V){
				if(x<0) S.erase(S.find(-x));
				else S.insert(x);
			}
			inc.erase(inc.begin());
		}
		if(S.empty()) cout<<-1<<'\n';
		else cout<<*S.begin()<<'\n';
		
	}

	return 0;
}
