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
// E - Roadwork
#include <bits/stdc++.h>

using namespace std;

int main(int argc, char *argv[])
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	int n, q, x;
	cin >> n >> q;
	vector<pair<int, pair<int, int> > > a(n);
	for (int i = 0; i < n; i++) cin >> a[i].second.first >> a[i].second.second >> a[i].first;
	set<pair<int, int> > d;
	for (int i = 0; i < q; i++) cin >> x, d.insert(make_pair(x, i));

	sort(a.begin(), a.end());

	vector<int> ans(q, -1);
	for (int i = 0; i < n; i++) {
		int l = a[i].second.first - a[i].first, r = a[i].second.second - a[i].first;
		set<pair<int, int> >::iterator it = d.lower_bound(make_pair(l, -1));
		while (d.size() > 0 && it != d.end()) {
			if (it->first >= r) break;
			ans[it->second] = a[i].first;
			d.erase(it++);
		}
	}

	for (int i = 0; i < q; i++) cout << ans[i] << endl;
	return 0;
}
