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;
typedef long long ll;
const ll INF = 1LL << 60;
const ll MOD = 1e9 + 7;

#define rep(i,n) for (int i = 0; i < (n); i++)
#define ALL(v) v.begin(), v.end()
typedef pair<int, int> P;
int X[200000];
int main() {
	int N, Q;
	cin >> N >> Q;
	vector<P> eve;
	for (int i = 0; i < N; i++) {
		int S, T;
		cin >> S >> T >> X[i];
		eve.emplace_back(max(S - X[i], 0), -i);
		eve.emplace_back(max(T - X[i], 0), -i);
	}
	for (int i = 1; i <= Q; i++) {
		int D;
		cin >> D;
		eve.emplace_back(D, i);
	}
	sort(ALL(eve));
	set<int> S;
	multiset<int> val;
	for (P p : eve) {
		if (p.second <= 0) {
			if (S.count(-p.second) == 0) {
				S.insert(-p.second);
				val.insert(X[-p.second]);
			}
			else {
				val.erase(val.find(X[-p.second]));
				S.erase(-p.second);
			}
		}
		else {
			if (val.empty()) cout << -1 << endl;
			else cout << *val.begin() << endl;
		}
	}
}