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>
#define pi pair<int, int>
using namespace std;

int main() 
{
	ios_base::sync_with_stdio(false);
	int n, q, x, y, z;
	cin >> n >> q;
	pair <int, pi> coord[n];
	for (int i = 0; i < n; i++)
	{
		cin >> x >> y >> z;
		coord[i] = make_pair(z, make_pair(x - z, y - z - 1));
	}
	sort(coord, coord + n);
	
	set <int> p;
	map <int, int> ans;
	for (int i = 1; i <= q; i++)
	{
		cin >> x;
		p.insert(p.end(), x);
		ans[x] = -1;
	}
	
	for (int i = 0; i < n; i++)
	{
		x = coord[i].first;
		int start = coord[i].second.first;
		int end = coord[i].second.second;
		while (p.lower_bound(start) != p.end() && *p.lower_bound(start) <= end)
		{
			ans[*p.lower_bound(start)] = x; 
			p.erase(p.lower_bound(start));
		}
	}
	
	for (auto i: ans)
		cout << i.second << endl;
}