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;

void solve(void)
{
	int n, q;
	cin >> q >> n;
	vector<tuple<int, int, int>> qs;
	multiset<int> xs;
	for (int i = 0; i < q; i++)
	{
		int s, t, x;
		cin >> s >> t >> x;
		qs.push_back(make_tuple(s-x, 0, x));
		qs.push_back(make_tuple(t-x, 1, x));
	}
	sort(qs.begin(), qs.end());
	auto it = qs.begin();

	while (n--)
	{
		int d;
		cin >> d;
		while(get<0>(*it) <= d && it != qs.end())
		{
			int x = get<2>(*it);
			if (get<1>(*it))
			{
				xs.erase(xs.find(x));
			}
			else
			{
				xs.insert(x);
			}
			it++;
		}
		if (xs.empty())
		{
			cout << -1 << endl;
		}
		else
		{
			cout << *xs.begin() << endl;
		}
	}
}

int main()
{
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}
