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 < (int)n; i++)
using ll = long long;
using P = pair<int,int>;

int main(){
  int n, q;
  cin >> n >> q;
  vector<P> v;
  int s[n], t[n], x[n];
  rep(i,n) {
    cin >> s[i] >> t[i] >> x[i];
    v.push_back(P((s[i]-x[i])*2-1,i+1));
    v.push_back(P((t[i]-x[i])*2-1,n+i+1));
  }
  int d[q];
  rep(i,q) {
    cin >> d[i];
    v.push_back(P(d[i]*2,-i));
  }
  sort(v.begin(),v.end());
  vector<int> ans(q,-1);
  multiset<int> event;
  for(auto k : v) {
    if(k.second > 0 && k.second <= n) {
      event.insert(x[k.second-1]);
    } else if(k.second > n) {
      event.erase(event.find(x[k.second-n-1]));
    } else {
      if(!event.empty()) ans[-k.second] = *event.begin();
    }
  }
  for(auto k : ans) cout << k << endl;
  return 0;
}