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 <utility>
#include <set>
#include <algorithm>
#define llint long long
#define mod 1000000007

using namespace std;
typedef pair<llint, llint> P;
typedef pair<P, llint> E;

llint n, Q;
vector<E> vec;
multiset<llint> S;

int main(void)
{
  ios::sync_with_stdio(0);
  cin.tie(0);

  cin >> n >> Q;
	llint s, t, x;
	for(int i = 1; i <= n; i++){
		cin >> s >> t >> x;
		t--;
		s-=x, t -= x;
		vec.push_back(E(P(s, 0LL), x));
		vec.push_back(E(P(t, 2LL), x));
	}
	for(int i = 1; i <= Q; i++){
		cin >> x;
		vec.push_back(E(P(x, 1LL), 0LL));
	}
	sort(vec.begin(), vec.end());

	for(int i = 0; i < vec.size(); i++){
		llint t = vec[i].first.second;
		if(t == 0) S.insert(vec[i].second);
		if(t == 2) S.erase(S.lower_bound(vec[i].second));
		if(t == 1){
			if(S.size() == 0) cout << -1 << endl;
			else cout << *S.begin() << endl;
		}
	}

  return 0;
}
