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
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
#define rep(i,a,b) for (int i = (a); i < (b); i++)
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef complex<double> com;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const int inf = 2e9;



int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, q; cin >> n >> q;
	vector<pair<int, P>> f(n);
	set<P> st;
	rep(i, 0, n) cin >> f[i].second.first >> f[i].second.second >> f[i].first;
	rep(i, 0, q) {
		int p; cin >> p;
		st.emplace(p, i);
	}
	sort(f.begin(), f.end());
	vector<int> ans(q, -1);
	rep(i, 0, n) {
		int s = f[i].second.first-f[i].first;
		int t = f[i].second.second - f[i].first;
		while (true) {
			auto idx = st.lower_bound({ s,-inf });
			if (idx == st.end() ||(*idx).first>=t) break;
			ans[(*idx).second] = f[i].first;
			st.erase(idx);
		}
	}
	rep(i, 0, q) printf("%d\n", ans[i]);
	return 0;
}
