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<vector>
#include<string>
#include<set>
#include<algorithm>
#include<math.h>

using namespace std;

using p = pair<int, long long>;
using event = pair<long long, p>;

int main(){
  int n, q;
  cin >> n >> q;
  vector<event> v;
  for(int i = 0; i < n ; i++){
    int a,b,c;
    cin >> a >> b >> c;
    v.push_back(event(a-c, p(-1, c)));
    v.push_back(event(b-c, p(0, c)));
  }
  for(int i = 0; i < q; i++){
    int d; cin >> d;
    v.push_back(event(d, p(1, i)));
  }
  sort(v.begin(), v.end());

  multiset<long long> s;
  vector<long long> res(q);
  for(auto t : v){
    int w = t.second.first;
    long long val = t.second.second;
    if(w == -1) s.insert(val);
    else if(w == 0) s.erase(s.lower_bound(val));
    else res[val] = (s.empty() ? -1 : *(s.begin()));
  }
  for(auto v : res) cout << v << endl;
}
