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>
using namespace std;

typedef long long ll;

ll N, Q;
map<ll, set<pair<bool, ll>>> Events;
map<ll, ll> Dist;
ll INF = 1e10;

int main() {
  cin >> N >> Q;
  for (ll i = 0; i < N; ++i) {
    ll s, t, x;
    cin >> s >> t >> x;
    Events[s - x].insert(make_pair(true, x));
    Events[t - x].insert(make_pair(false, x));
  }
  multiset<ll> xs;
  xs.insert(INF);
  for (auto t_events : Events) {
    ll t = t_events.first;
    for (auto plus_x : t_events.second) {
      bool plus = plus_x.first;
      ll x = plus_x.second;
      if (plus)
        xs.insert(x);
      else
        xs.erase(xs.find(x));
    }
    Dist[t] = *xs.begin();
  }
  for (ll i = 0; i < Q; ++i) {
    ll d;
    cin >> d;
    auto it = Dist.upper_bound(d);
    ll res = it == Dist.begin() ? INF : prev(it)->second;
    cout << (res == INF ? -1 : res) << endl;
  }
  return 0;
}
