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;

struct Plan {
  int s, t, x;
  bool operator < (const Plan &p) const {
    return x < p.x;
  }
};

int main() {
  int n, q;
  cin >> n >> q;
  Plan p[n];
  rep(i, n) cin >> p[i].s >> p[i].t >> p[i].x;
  sort(p, p+n);

  int d[q];
  for (int i = q-1; i >= 0; --i) cin >> d[i];
  rep(i, q) d[i] *= -1;

  set<int> sets;
  rep(i, q) sets.insert(i);
  int ans[q];
  rep(i, q) ans[i] = 0;
  
  rep(i, n) {
    int l = distance(d, upper_bound(d, d+q, p[i].x-p[i].t));
    int r = distance(d, upper_bound(d, d+q, p[i].x-p[i].s));
    auto itrl = sets.lower_bound(l);
    auto itrr = sets.lower_bound(r);
    while (itrl != itrr) {
      ans[*itrl] = p[i].x;
      itrl = sets.erase(itrl);
    }
  }

  for (int i = q-1; i >= 0; --i) {
    cout << (ans[i] ? ans[i] : -1) << endl;
  }
  return 0;
}