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(){
	vector<tuple<int,int,int> > bye, bys;
	int n, q; cin >> n >> q;
	for(int i = 0; i < n; i++){
		int s, t, x; cin >> s >> t >> x;
		t--;
		bye.push_back(make_tuple(s - x, t - x, x));
		bys.push_back(make_tuple(t - x, s - x, x));
	}
	sort(bye.begin(), bye.end());
	sort(bys.begin(), bys.end());
	multiset<int> now;
	int add = 0, rem = 0;
	for(int i = 0; i < q; i++){
		int d; cin >> d;
		while(add < n){
			int s, t, x; tie(s, t, x) = bye[add];
			if(s <= d){
				now.insert(x);
				add++;
			} else {
				break;
			}
		}
		while(rem < n){
			int s, t, x; tie(t, s, x) = bys[rem];
			if(t < d){
				now.erase(now.find(x));
				rem++;
			} else {
				break;
			}
		}
		if(now.empty()) cout << -1 << endl;
		else cout << *(now.begin()) << endl;
	}
	return 0;
}
