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 r(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef pair<int,int>P;
typedef pair<int,P>P2;
#define F first
#define S second

signed main(){

  int n,q;

  priority_queue<P2,vector<P2>,greater<P2> >ib;

  cin>>n>>q;
  vector<int>Z;
  vector<P2>v;
  r(i,n){
    int s,t,x;
    scanf("%d%d%d",&s,&t,&x);
    s-=x;
    t-=x;
    s=max(0,s);
    if(t<0)continue;
    ib.push(P2(s,P(1,x)));
    ib.push(P2(t,P(-1,x)));
  }

  set<int>st;

  for(int i=0;i<q;i++){
    int x;
    cin>>x;
    while(!ib.empty() && ib.top().F<=x){
      P2 p=ib.top();ib.pop();
      if(p.S.F==1)st.insert(p.S.S);
      else st.erase(p.S.S);
    }
    if(st.size())cout<<*st.begin()<<endl;
    else cout<<-1<<endl;
  }


}