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 N, Q;

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> Q;
	vector<pair<int, int>> evs;
	for (int i = 0; i < N; i++) {
		int s, t, x; cin >> s >> t >> x;
		evs.emplace_back(s-x, x);
		evs.emplace_back(t-x, ~x);
	}
	sort(evs.begin(), evs.end());
	multiset<int> cnds;
	for (int q = 0, i = 0; q < Q; q++) {
		int d; cin >> d;
		while (i < int(evs.size()) && evs[i].first <= d) {
			if (evs[i].second >= 0) {
				cnds.insert(evs[i].second);
			} else {
				cnds.erase(cnds.find(~evs[i].second));
			}
			i++;
		}
		if (cnds.empty()) {
			cout << -1 << '\n';
		} else {
			cout << *cnds.begin() << '\n';
		}
	}

	return 0;
}