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 main() {
	int N, Q;
	cin >> N >> Q;
	vector<tuple<int, int, int>>stx(N);
	for (int n = 0;n<N;++n) {
		cin >> get<1>(stx[n]) >> get<2>(stx[n]) >> get<0>(stx[n]);
	}
	sort(stx.begin(), stx.end());

	set<pair<int, int>> d;
	for (int i = 0, x; i < Q; ++i) {
		cin >> x;
		d.emplace(x, i);
	}

	vector<long long>tm(Q, -1);
	for (int i = 0;i<N;++i) {
		auto first = d.lower_bound(make_pair(get<1>(stx[i]) - get<0>(stx[i]), 0));
		auto end = d.upper_bound(make_pair(get<2>(stx[i]) - get<0>(stx[i]) - 1, Q));

		for (auto it = first; it != end; ++it) {
			tm[it->second] = get<0>(stx[i]);
		}
		d.erase(first, end);
	}
	for (int i = 0; i < Q; ++i) cout << tm[i] << '\n';
	return 0;
}