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;
const int N = 1e6+100;
typedef long long ll;
vector<pair<int,pair<int,int> > > roadwork;
set<pair<ll,ll> > peo;
ll ans[N];
int main(){
	ll n,q;
	ll s,t,x,d;
	cin>>n>>q;
	for(ll i=1;i<=n;i++){
		cin>>s>>t>>x;
		roadwork.push_back({x,{s-x,t-1-x}});
	}
	for(ll i=1;i<=q;i++){
		cin>>d;
		peo.insert({d,i});
		ans[i]=-1;
	}
	sort(roadwork.begin(),roadwork.end());
	for(auto work:roadwork){
		ll L=work.second.first;
		ll R=work.second.second;
		if(peo.empty()) continue;
		auto it=peo.lower_bound({L,-1});
		while(!peo.empty()&&it!=peo.end()&&(*it).first<=R){
			ans[(*it).second]=work.first;
			peo.erase(it++);
		}
	} 
	for(ll i=1;i<=q;i++){
		cout<<ans[i]<<endl;
	}
	return 0;
}
