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<vector>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
int N, Q;
const int NQmax = 2e5;
ll S[NQmax], T[NQmax], X[NQmax], D[NQmax];

int main(){
	cin >> N >> Q;
	for(int i=0; i<N; i++) cin >> S[i] >> T[i] >> X[i];
	for(int i=0; i<Q; i++) cin >> D[i];

	vector<pair<int, int> > v;
	for(int i=0; i<N; i++){
		v.push_back({S[i]-X[i], i});
	}
	for(int i=0; i<Q; i++){
		v.push_back({D[i], i+N});
	}
	sort(v.begin(), v.end());

	priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > que;
	ll ans = 0;
	for(int i=0; i<N+Q; i++){
		int x = v[i].first;
		int j = v[i].second;
		if(j < N){
			que.push({X[j], j});
		} else {
			j -= N;
			while(!que.empty() && T[que.top().second]-X[que.top().second]<=x) que.pop();
			if(que.empty()){
				cout << -1 << endl;
			} else {
				cout << que.top().first << endl;
			}
		}
	}
	return 0;
}
