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 <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>

using namespace std;

typedef pair<int, int> P;
typedef pair<P, int> PP;

int n, q;
priority_queue<PP, vector<PP>, greater<PP> > pq;
set<int> st;

int main() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    int s, t, x;
    cin >> s >> t >> x;
    pq.push(PP(P(s-x, 1), x));
    pq.push(PP(P(t-x, -1), x));
  }
 
  while(q--) {
    int d;
    cin >> d;
    while (pq.size() and pq.top().first.first <= d) {
      if (pq.top().first.second == 1) {
        st.insert(pq.top().second);
      } else {
        st.erase(pq.top().second);
      }
      pq.pop();
    }
    if (st.empty()) {
      cout << -1 << endl;
      continue;
    }
    cout << *st.begin() << endl;
  }
  
  return 0;
}