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;

const int N = 2e5 + 100;

int n, q, l[N], r[N], x[N], d[N], ans[N];
vector<int> add[N], del[N];// avval add bad del
multiset<int> st;

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> q;
	for (int i = 0; i < n; i++) {
		cin >> l[i] >> r[i] >> x[i];
		l[i] = max(0, l[i] - x[i]);
		r[i] = max(0, r[i] - x[i]);
	}
	for (int i = 0; i < q; i++)
		cin >> d[i];
	for (int i = 0; i < n; i++) {
		add[lower_bound(d, d + q, l[i]) - d].push_back(x[i]);
		del[lower_bound(d, d + q, r[i]) - d].push_back(x[i]);
	}
	for (int i = 0; i < q; i++) {
		for (int num: add[i])
			st.insert(num);
		for (int num: del[i])
			st.erase(st.find(num));
		ans[i] = (st.empty()? -1: *st.begin());
	}
	for (int i = 0; i < q; i++)
		cout << ans[i] << '\n';
	return 0;
}