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;
#define INF 1LL<<62
#define inf 1000000007 
	ll n,q,ans[200010];
	set<pair<ll,ll>>d;
	vector<tuple<ll,ll,ll>>ch;
int main() {

	cin>>n>>q;
	for(ll i=0;i<n;i++){
		ll S,T,X;
		cin>>S>>T>>X;
		T--;
		S-=X;
		T-=X;
		ch.push_back(make_tuple(X,S,T));
	}
	sort(ch.begin(),ch.end());
	for(ll i=0;i<q;i++){
		ll now;
		cin>>now;
		d.insert(make_pair(now,i));
		ans[i]=-1;
	}
	for(ll i=0;i<n;i++){
		ll now1=get<1>(ch[i]);
		ll now2=get<2>(ch[i]);
		ll now3=get<0>(ch[i]);
		auto itr=d.lower_bound(make_pair(now1,-1));		
		while(itr!=d.end()){
			if(now2>=itr->first){
			ans[itr->second]=now3;
			itr=d.erase(itr);
			}
			else{
				break;
			}
		}

	}
	for(ll i=0;i<q;i++){
		cout <<ans[i]<<endl;
	}
	// your code goes here
	return 0;
}