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;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,ll> type;// 1: query, 0: end, -1: begin
typedef pair<ll,type> event;

int main(){
  int N,Q;
  cin >> N >> Q;
  vector<event> v;
  rep(i,N) {
    int s,t,x;
    cin >> s >> t >> x;
    v.push_back({s-x, type(-1,x)});
    v.push_back({t-x, type(0,x)});
  }
  rep(i,Q) {
    int d;
    cin >> d;
    v.push_back({d, type(1,i)});
  }
  sort(v.begin(), v.end());

  vector<ll> res(Q);
  multiset<ll> se;

  for (auto p : v) {
    ll x = p.first;
    int type = p.second.first;
    ll val = p.second.second;
    if (type == -1) se.insert(val);
    else if (type == 0) se.erase(se.lower_bound(val));
    else res[val] = (se.empty() ? -1 : *(se.begin()));
  }
  for (auto e : res) cout << e << endl;
  return 0;
}