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 ll = long long;
using P = pair<ll ,ll>;
using T = tuple<ll, ll, ll>;
const ll INF = 1LL<<61LL;
const ll MOD = 1e9+7;

int main(){
	ll N, Q;
	cin >> N >> Q;
	vector<T> stop(N);
	for(int i=0;i<N;i++){
		ll s, t, x;
		cin >> s >> t >> x;
		t--;
		stop[i] = T(x, s, t);
	}
	sort(stop.begin(), stop.end());
	set<P> st;
	for(int i=0;i<Q;i++){
		ll q;
		cin >> q;
		st.insert(P(q, i));
	}
	vector<ll> ans(Q, -1);
	for(auto &&quit : stop){
		ll x, s, t;
		tie(x, s, t) = quit;
		auto ite = st.lower_bound(P(s-x, -10));
		while(ite != st.end() && ite->first <= t-x){
			auto tmp = ite;
			ans[ite->second] = x;
			ite++;
			st.erase(tmp);
		}
	}
	for(int i=0;i<Q;i++)
		cout << ans[i] << endl;
	return 0;
}
