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 <set>
#include <utility>
#include <algorithm>
using namespace std;
pair<int,pair<int,int>> p[200010];
set<pair<int,int>> s;
int ans[200010];
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int i,n,q;
	cin >> n >> q;
	for(i=0;i<n;i++){
		int l,r,x;
		cin >> l >> r >> x;
		p[i] = {x,{l,r}};
	}
	sort(p,p + n);
	for(i=0;i<q;i++){
		int d; cin >> d;
		s.insert({d,i}); ans[i] = -1;
	}
	for(i=0;i<n;i++){
		int x = p[i].first;
		int l = p[i].second.first - x,r = p[i].second.second - x;
		auto it = s.lower_bound({l,-1});
		while(it!=s.end()){
			if(it->first>=r) break;
			ans[it->second] = x;
			s.erase(it++);
		}
	}
	for(i=0;i<q;i++){
		cout << ans[i] << "\n";
	}
}