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;

int main() {
  int N,Q;
  cin >> N >> Q;

  vector< tuple< int, int, int > > stx(N);
  for (int i=0;i<N; ++i) {
    int sin, tin, xin;
    cin >> sin >> tin >> xin;
    stx[i] = make_tuple( xin, sin-xin, tin-xin );
  }

  sort( stx.begin(), stx.end() );

  set< pair< int,int > > D;
  vector< int > DO(Q,-1);
  
  for (int i=0; i<Q; ++i ) {
    int din;
    cin >> din;
    D.emplace( din,i );
  }
  for ( auto a: stx ) {
//      cout << "#" << get<0>(a) << " " << get<1>(a) << " " << get<2>(a) << endl;
    auto first = D.lower_bound( make_pair( get<1>(a), 0) );
    auto last = D.upper_bound( make_pair( get<2>(a)-1, Q) );
    for ( auto it = first; it != last; ++it ) {
      DO[ it->second ] = get<0>(a);
    }
    D.erase(first,last);
  }
  for (int i=0;i<Q;++i ) {
    cout << DO[i] << endl;
  }
}

