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;

#define sz(a) int(a.size())

int main() {
	int n, q;
	cin >> n >> q;

	map <int, vector <int> > ev_I, ev_O;

	set <int> ev;
	for (int i = 0; i < n; i++) {
		int s, t, x;
		cin >> s >> t >> x;
		ev_I[s-x].push_back(x);
		ev_O[t-x].push_back(x);
		ev.insert(s-x), ev.insert(t-x);
	}

	multiset<int> cur;
	int last = -1;

	vector <int> D (q);
	for (int &t : D) cin >> t;

	int id = 0;
	for (auto t : ev) {
		if (ev_O.count(t)) {
			for (auto r : ev_O[t]) cur.erase(cur.find(r));
		}
		if (ev_I.count(t)) {
			for (auto r : ev_I[t]) cur.insert(r);
		}

		while (id < q && D[id] < t) id++, cout << last << endl;

		last = cur.empty() ? -1 : *begin(cur);
	}

	while (id < q) id++, cout << -1 << endl;


}

