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 <cstdio>
#include <set>
#include <vector>
#include <algorithm>

using namespace std;

struct Roadwork {
	int kin, dis, tim;
	bool operator < (Roadwork other) const {
		return tim > other.tim;
	}
};

int n, q;
vector<Roadwork> road;

int main(int argc, char const *argv[]) {
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; ++i) {
		int s, t, x;
		scanf("%d%d%d", &s, &t, &x);
		road.push_back((Roadwork){ 1, x, s - x});
		road.push_back((Roadwork){ 0, x, t - x});
	}
	sort(road.begin(), road.end());
	multiset<int> ms;
	for (int i = 1; i <= q; ++i) {
		int d;
		scanf("%d", &d);
		while (!road.empty() && road.back().tim <= d) {
			if (road.back().kin == 1) {
				ms.insert(road.back().dis);
			} else {
				ms.erase(ms.find(road.back().dis));
			}
			road.pop_back();
		}
		printf("%d\n", ms.empty() ? -1 : *ms.begin());
	}

	return 0;
}
