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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const ll MAX_N=2e5,MAX_Q=2e5;
ll N,Q,S[MAX_N],T[MAX_N],X[MAX_N],ans[MAX_Q];
typedef pair<ll,ll> P;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N>>Q;
	REP(i,N)cin>>S[i]>>T[i]>>X[i];
	set<P> st;
	REP(i,Q){
		ll d;
		cin>>d;
		st.insert(P(-d,i));
	}
	vector<ll> stop;
	REP(i,N)stop.push_back(i);
	sort(stop.begin(),stop.end(),[](ll a,ll b){
		return X[a]<X[b];
	});
	REP(i,Q)ans[i]=-1;
	for(auto i:stop){
		ll s=S[i],t=T[i],x=X[i];
		auto ite=st.upper_bound(P(x-t,N));
		while(ite!=st.end()&&ite->first<=x-s){
			ans[ite->second]=x;
			ite=st.erase(ite);
		}
	}
	REP(i,Q)cout<<ans[i]<<endl;
}