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;

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

	ll N,Q;
	cin >> N >> Q;
	vector<pair<ll,ll>> x(N);
	vector<ll> s(N),t(N);
	for(int i=0;i<N;++i){
		cin >> s[i] >> t[i] >> x[i].first;
		x[i].second = i;
		s[i] -= x[i].first;
		t[i] -= x[i].first;
	}
	sort(x.begin(),x.end());
	set<ll> d;
	map<ll,ll> ma;
	for(int i=0;i<Q;++i){
		ll di;
		cin >> di;
		ma[di] = i;
		d.insert(di);
	}
	vector<ll> ans(Q,-1);
	for(int i=0;i<N;++i){
		ll xi = x[i].first;
		ll si = s[x[i].second];
		ll ti = t[x[i].second];
		vector<ll> del;
		auto l = d.lower_bound(si);
		auto r = d.lower_bound(ti);
		while(l!=r){
			ans[ma[*l]] = xi;
			del.push_back(*l);
			l++;
		}
		for(auto er:del) d.erase(er);
	}
	for(auto a:ans){
		cout << a << endl;
	}
}
