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 (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;


int main()
{
  int n, q;
  cin >> n >> q;
  priority_queue<tuple<int, bool, int>, vector<tuple<int, bool, int>>, greater<tuple<int, bool, int>>> event;
  rep(i, n)
  {
    int s, t, x;
    cin >> s >> t >> x;
    event.emplace(s - x, true, x);
    event.emplace(t - x, false, x);
  }
  vector<int> d(q);
  rep(i, q)
    cin >> d[i];
  sort(d.begin(), d.end());
  set<int> x;
  rep(i, q)
  {
    while (!event.empty() && get<0>(event.top()) <= d[i])
    {
      if (get<1>(event.top()))
        x.insert(get<2>(event.top()));
      else
        x.erase(get<2>(event.top()));
      event.pop();
    }
    if (x.empty())
      printf("-1\n");
    else
      printf("%d\n", *x.begin());
  }
  return 0;
}