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 <iostream>
#include <set>
#include <tuple>
#include <algorithm>

using namespace std;

int main(int argc, char *argv[])
{
	int N, Q;
	tuple<int, int, int> stx[200001];
	set<pair<int, int> > st;

	cin >> N >> Q;

	for (int i = 0; i < N; i++) {
		int S, T, X;
		cin >> S >> T >> X;
		stx[i] = make_tuple(X, S - X, T - X);
	}
	sort(stx, stx + N);

	for (int i = 0; i < Q; i++) {
		int D;
		cin >> D;
		st.insert(make_pair(D, i));
	}

	int ans[200001] = {0};
	for (int i = 0; i < Q; i++) {
		ans[i] = -1;
	}
	for (int i = 0; i < N; i++) {
		int S, T, X;
		tie(X, S, T) = stx[i];

		auto itr = st.lower_bound(make_pair(S, -1));
		while (itr != st.end()) {
			if (itr->first >= T) {
				break;
			}
			ans[itr->second] = X;
			st.erase(itr++);
		}
	}

	for (int i = 0; i < Q; i++) {
		cout << ans[i] << "\n";
	}

	return 0;
}
