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>
#define pb push_back
#define ll long long
#define mp make_pair
#define inf 1000000007
using namespace std;
int main(){
	int n,q;
	cin>>n>>q;
	vector<int> s,t,x;
	vector< pair<int,pair<int,int> > > ev;
	for(int i=0;i<n;i++){
		int ts,tt,tx;
		cin>>ts>>tt>>tx;
		s.pb(ts-tx);t.pb(tt-tx-1);x.pb(tx);
	}
	for(int i=0;i<n;i++){
		ev.pb( mp(s[i],mp(-1,x[i])) );
		ev.pb( mp(t[i],mp(1,x[i])) );
	}
	for(int i=0;i<q;i++){
		int tm;
		cin>>tm;
		ev.pb( mp(tm, mp(0,0)) );
	}
	sort( ev.begin(), ev.end() );
	set<int> fusa;
	for(int i=0;i<ev.size();i++){
		int tim = ev[i].first;
		int tp = ev[i].second.first;
		int pl = ev[i].second.second;
		//cout<<tim<<" "<<tp<<" "<<pl<<endl;
		if(tp == -1){
			fusa.insert(pl);
		}
		if(tp == 0){
			if(fusa.empty()){
				cout<<-1<<endl;
			}
			else{
				cout<<*fusa.lower_bound(-1)<<endl;
			}
		}
		if(tp == 1){
			fusa.erase(pl);
		}
	}
	return 0;
}