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 <vector>
#include <tuple>
#include <algorithm>
#include <set>

using namespace std;


int main()
{
	int N, Q;
	cin >> N >> Q;

	vector<tuple<int, int, int>> X;

	for (int i = 0; i < N; i++)
	{
		int s, t, x;
		cin >> s >> t >> x;

		X.push_back(make_tuple(x, s, t));
	}

	set<pair<int, int>> D;

	for (int i = 0; i < Q; i++)
	{
		int d;
		cin >> d;
		D.insert(make_pair(d, i));
	}

	sort(X.begin(), X.end());
	
	vector<int> kekka(Q, -1);

	for (int i = 0; i < N; i++)
	{
		int s, t, x;

		tie(x, s, t) = X[i];

		auto it = D.lower_bound(make_pair(s - x, -1));

		while (it != D.end())
		{
			if(it->first >= t -x)
			{
				break;
			}
			kekka[it->second] = x;
			D.erase(it++);
		}
	}

	for (int i = 0; i < Q; i++)
	{
		cout << kekka[i] << endl;
	}

	return 0;
}
