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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<pair<int, int>> s, t;
  for (int i = 0; i < n; i++) {
    int ss, tt, x;
    cin >> ss >> tt >> x;
    s.emplace_back(ss - x, x);
    t.emplace_back(tt - x, x);
  }
  sort(s.begin(), s.end());
  sort(t.begin(), t.end());
  vector<int> d(q);
  for (int i = 0; i < q; i++) {
    cin >> d[i];
  }
  map<int, int> st;
  int c = 0, cc = 0;
  for (int i = 0; i < q; i++) {
    for (; c < n && s[c].first <= d[i]; c++) {
      st[s[c].second]++;
    }
    for (; cc < n && t[cc].first <= d[i]; cc++) {
      if (st[t[cc].second] == 1) st.erase(t[cc].second);
      else st[t[cc].second]--;
    }
    if (st.empty()) {
      cout << -1 << endl;
    } else {
      cout << st.begin()->first << endl;
    }
  }
  return 0;
}