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;

using Pi = pair < int , int >;

int N, Q;
int S[200005], T[200005], X[200005];
vector < Pi > push[200005], pop[200005];
int D[200005];

int main()
{

	cin >> N >> Q;
	for ( int i = 0; i < N; i++ ) {
		cin >> S[i] >> T[i] >> X[i];
		S[i] -= X[i], T[i] -= X[i];
	}
	
	for ( int i = 0; i < Q; i++ ) {
		cin >> D[i];
	}
	
	for ( int i = 0; i < N; i++ ) {
		int ps = lower_bound(D, D + Q, S[i]) - D;
		int pt = lower_bound(D, D + Q, T[i]) - D;
		push[ps].push_back(Pi(X[i], i));
		pop[pt].push_back(Pi(X[i], i));
	}
	
	set < Pi > st;
	for ( int i = 0; i < Q; i++ ) {
		for ( auto &x : push[i] ) st.insert(x);
		for ( auto &x : pop[i] ) st.erase(x);
		if ( !st.empty() ) {
			cout << (st.begin())->first << endl;
		} else {
			cout << -1 << endl;
		}
	}

	return ( 0 );
}
