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 <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;

int main(){
  int N,Q;
  cin>>N>>Q;
  vector<pair<lint,pair<int,int> > > m;
  REP(i,N){
    int S,T,X;
    cin>>S>>T>>X;
    m.push_back(make_pair(S-X,make_pair(X,1)));
    m.push_back(make_pair(T-X,make_pair(X,-1)));
  }
  sort(ALL(m));
  set<lint> s;
  lint D[Q];
  REP(i,Q){
    cin>>D[i];
  }
  auto itr=m.begin();
  REP(i,Q){
    while(itr->first<=D[i] and itr!=m.end()){
      if(itr->second.second==1)s.insert(itr->second.first);
      else s.erase(itr->second.first);
      itr++;
    }
    if(s.empty())cout<<-1<<endl;
    else cout<<*s.begin()<<endl;
  }

  return 0;
}
