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;

int main() {
	int N, Q;

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

	vector<tuple<int, int, int>> v;
	for( int i = 0; i < N; i++ ) {
		v.push_back( make_tuple( S[i] - X[i], 1, X[i] ) );
		v.push_back( make_tuple( T[i] - X[i], -1, X[i] ) );
	}
	sort( v.begin(), v.end() );
	int j = 0;
	set<int> a;
	for( int i = 0; i < Q; i++ ) {
		while( j < v.size() && get<0>( v[j] ) <= D[i] ) {
			if( get<1>( v[j] ) > 0 ) {
				a.insert( get<2>( v[j] ) );
			}
			else {
				a.erase( get<2>( v[j] ) );
			}
			j++;
		}
		int ans = -1;
		if( !a.empty() ) {
			ans = *a.begin();
		}
		cout << ans << endl;
	}

	return 0;
}
