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 <vector>
#include <tuple>
#include <set>
#include <algorithm>
using namespace std;

int main() {
	int N, Q;
	scanf("%d %d", &N, &Q);
	vector<tuple<int, int, int>> event;
	for(int i = 0; i < N; i++) {
		int s, t, x;
		scanf("%d %d %d", &s, &t, &x);
		event.push_back(make_tuple(s-x, 1, x));
		event.push_back(make_tuple(t-x, -1, x));
	}
	
	vector<int> D(Q);
	for(int i = 0; i < Q; i++) scanf("%d", &D[i]);
	
	sort(event.begin(), event.end());
	multiset<int> st;
	int k = 0;  //何番目のイベントまで処理したか
	for(int i = 0; i < Q; i++) {
		while(k < event.size() && get<0>(event[k]) <= D[i]) {
			if(get<1>(event[k]) == 1) {
				st.insert(get<2>(event[k]));
			} else {
				auto it = st.find(get<2>(event[k]));
				if(it != st.end()) st.erase(it); //1つだけ削除
			}
			k++;
		}
		
		if(st.empty()) printf("-1\n");
		else printf("%d\n", *st.begin());
	}
	
	return 0;
}