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 <stdio.h>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, l, n, q;
  scanf("%lld%lld", &n, &q);
  vector<ll> vec[q + 1];
  ll s[n], t[n], x[n], d[q];
  for(i = 0; i < n; ++i) scanf("%lld%lld%lld", &s[i], &t[i], &x[i]);
  for(i = 0; i < q; ++i) scanf("%lld", &d[i]);
  for(i = 0; i < n; ++i) {
    vec[lower_bound(d, d + q, s[i] - x[i]) - d].push_back(x[i]);
    vec[lower_bound(d, d + q, t[i] - x[i]) - d].push_back(-x[i]);
  }
  multiset<ll> ans;
  for(i = 0; i < q; ++i) {
    for(j = 0; j < vec[i].size(); ++j) {
      if(vec[i][j] > 0) ans.insert(vec[i][j]);
      else ans.erase(ans.find(-vec[i][j]));
    }
    if(ans.size()) printf("%lld\n", (*ans.begin()));
    else printf("-1\n");
  }
  return 0;
}