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 ll long long int
#define MOD 1000000007
#define INF 1e18
#define PI 3.14159265358979

using namespace std;

int main(void){
  ll n, q;
  cin >> n >> q;

  vector<tuple<ll,ll,ll,ll>> ev;
  for (int i = 0; i < n; i++){
    ll s, t, x;
    cin >> s >> t >> x;
    
    ev.emplace_back(s-x, 0, x, 1);
    ev.emplace_back(t-x, 0, x, -1);
  }
  for (ll i = 0; i < q; i++){
    ll d;
    cin >> d;
    
    ev.emplace_back(d, 1, i, 0);
  }

  vector<ll> ans(q);
  multiset<ll> s;

  sort(ev.begin(), ev.end());

  
  for (auto e : ev){
    ll com = get<3>(e);
    ll val = get<2>(e);
    if (com == 1){
      s.insert(val);
    }
    else if (com == -1){
      s.erase(s.lower_bound(val));
    }
    else {
      ans[val] = (s.empty() ? -1 : *(s.begin()));
    }
  }

  for (auto v : ans){
    cout << v << endl;
  }
  
  


  return 0;
}
