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;
using llong = long long;
using ldbl = long double;
using P = pair<llong, llong>;
using PP = pair<P, llong>;

#define ALL(x) x.begin(), x.end()

const llong inf = 1100100100100100ll;
const llong mod = 1e9+7;

int main() {
	llong N, Q;
	vector<PP> event;

	cin >> N >> Q;
	event.resize(N*2 + Q);

	llong S, T, X;
	for (int i = 0; i < N; i++) {
		cin >> S >> T >> X;
		event[i] = PP(P(S-X,1), X);
		event[i+N] = PP(P(T-X,0), X);
	}
	llong D;
	for (int i = 0; i < Q; i++) {
		cin >> D;
		event[N*2 + i] = PP(P(D,2), 0);
	}

	sort(ALL(event));
	set<llong> s;
	for (auto eve : event) {
		int tm = eve.first.first;
		int id = eve.first.second;
		int pos = eve.second;

		if (id == 0) {
			s.erase(pos);
		} else if (id == 1) {
			s.insert(pos);
		} else {
			if (s.size() == 0)
				cout << -1 << endl;
			else
				cout << *s.begin() << endl;
		}
	}

	return 0;
}