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 rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=1e18;




int main() {
  ll n,q;
  cin >> n >> q;
  vector<tuple<ll,ll,ll>> a(n);
  rep(i,n){
    ll s1,t1,x1;
    cin>>s1>>t1>>x1;
    a[i]=make_tuple(x1,s1-x1,t1-x1);
  }
  sort(a.begin(),a.end());
  vector<ll>ans(q,-1);
  set<P> Q;
  rep(i,q){
    ll d1;
    cin>>d1;
    Q.insert(make_pair(d1,i));
  }
rep(i,n){
  ll s1,t1,x1;
  tie(x1,s1,t1)=a[i];
  auto it=Q.lower_bound(make_pair(s1,ll(-1)));
  while(it!=Q.end()){
    if((it->first)>=t1)break;
    ans[it->second]=x1;
  //  cout<<it->second<<" "
  //  if()
    Q.erase(it++);
  }
}
rep(i,q){
  cout<<ans[i]<<endl;
}


}
