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 <stdio.h>
#include <algorithm>
#include <queue>
#include <set>
#include <functional>
using namespace std;
using ll = long long;

using pii = pair<int, int>;
int n, q, s[200010], t[200010], x[200010], d[200010];
set<int> block;
priority_queue<pii, vector<pii>, greater<pii>> st, ed;
int main() {
	scanf("%d %d", &n, &q);
	for (int i = 0; i < n; i++) {
		scanf("%d %d %d", s + i, t + i, x + i);
		s[i] -= x[i];
		t[i] -= x[i];
		st.emplace(s[i], i);
	}
	for (int i = 0; i < q; i++)
		scanf("%d", d + i);
	for (int i = 0; i < q; i++) {
		while (!ed.empty() && ed.top().first <= d[i]) {
			block.erase(x[ed.top().second]);
			ed.pop();
		}
		while (!st.empty() && st.top().first <= d[i]) {
			int id = st.top().second;
			if (d[i] < t[id]) {
				block.emplace(x[id]);
				ed.emplace(t[id], id);
			}
			st.pop();
		}
		if (block.empty())
			printf("-1\n");
		else
			printf("%d\n", *block.begin());
	}
}