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 + 3;
int res[N];
int main() {
#ifndef ONLINE_JUDGE
//	freopen("input.in", "r", stdin);
//	freopen("output.out", "w", stdout);
#endif
	int n, q;
	scanf("%d %d", &n, &q);
	vector<pair<int, pair<int, int>>> events;
	for (int i = 0; i < n; ++i) {
		int s, e, x;
		scanf("%d %d %d", &s, &e, &x);
		events.push_back({x, {s, e - 1}});
	}
	set<pair<int, int>> st;
	for (int i = 0; i < q; ++i) {
		int s;
		scanf("%d", &s);
		st.insert({s, i});
	}
	sort(events.begin(), events.end());
	memset(res, -1, sizeof res);
	for (auto event: events) {
		int l = event.second.first - event.first;
		int r = event.second.second - event.first;
		auto it = st.lower_bound({l, 0});
		while (it != st.end() && it->first <= r) {
			res[it->second] = event.first;
			it = st.erase(it);
		}
	}
	for (int i = 0; i < q; ++i)
		printf("%d\n", res[i]);
	return 0;
}
