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;

long long N, Q;
//X, <s-x, t-x>. Xが小さい方が先に当たる
vector<pair<long long, pair<long long, long long> > > XST;
set<pair<long long, long long> > DI;
long long ans[200005];

int main(){
	cin >> N >> Q;
	for(long long i = 0; i < N; i++){
		long long s,t,x;
		cin >> s >> t >> x;
		XST.push_back(make_pair(x, make_pair(s-x, t-x)));
	}
	sort(XST.begin(), XST.end());

	for(long long i = 0; i < Q; i++){
		long long d;
		cin >> d;
		DI.insert(make_pair(d, i));
	}

	for(long long i = 0; i < N; i++){
		long long x = XST[i].first;
		long long start = XST[i].second.first;
		long long end = XST[i].second.second;
		auto it = DI.lower_bound(make_pair(start, -1));
		while(it != DI.end() && (*it).first < end){
			ans[(*it).second] = x;
			DI.erase(it++);
		}
	}

	for(long long i = 0; i < Q; i++){
		if(ans[i] != 0){
		cout << ans[i] << endl;
		}else{
			cout << -1 << endl;
		}
	}
}
