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 <algorithm>
#include <vector>
#include <limits.h>
#include<set>
using namespace std;

#define rip(i,n,s) for (int i=(s);i<(int)(n);i++)

int main() {
	int n,q;
	cin >> n >> q;
	int s, t, x;
	vector<pair<int, int>> da(n * 2 + q);
	rip(i, n, 0) {
		cin >> s >> t >> x;
		da[i * 2].first = s - x;
		da[i * 2].second = x;
		da[i * 2 + 1].first = t - x;
		da[i * 2 + 1].second = -x;
	}
	int d;
	rip(i, q, 0) {
		cin >> d;
		da[n * 2 + i].first = d;
		da[n * 2 + i].second = INT_MAX;
	}
	set<int> se;
	sort(da.begin(), da.end());
	vector<int> ans(q);
	int co = 0;
	rip(i, n * 2 + q, 0) {
		if (da[i].second == INT_MAX) {
			if (se.empty()) {
				ans[co] = -1;
			}
			else {
				ans[co] = *se.begin();
			}
			co++;
		}
		else {
			if (da[i].second < 0) {
				se.erase(-da[i].second);
			}
			else {
				se.insert(da[i].second);
			}
		}
	}
	rip(i, q, 0) {
		printf("%d\n", ans[i]);
	}
}
