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 int ll;
#define rep(i,n) for(int i = 0; i < n; i++)

struct info {
  ll left,right,time;
};
 
bool comp(const info& a, const info& b) {
  return a.time < b.time;
}

int main() {
  int n,q;
  cin >> n >> q;
  ll s,t,x;
  vector<info> vec(n);
  rep(i,n) {
    cin >> s >> t >> x;
    vec[i] = {x-t+1,x-s,x};
  }
  sort(vec.begin(),vec.end(),comp);
  vector<ll> ans(q,-1);
  map<ll,int> mp;
  set<ll> Set;
  ll d;
  rep(i,q) {
    cin >> d;
    d = -d;
    mp[d] = i;
    Set.insert(d);
  }
  rep(i,n) {
    while (Set.size() > 0) {
      auto itr1 = Set.lower_bound(vec[i].left);
      if (itr1 != Set.end() && (*itr1) <= vec[i].right) {
        ans[mp[(*itr1)]] = vec[i].time;
        Set.erase((*itr1));
      }
      else break;
    }
  }
  rep(i,q) printf("%lld\n",ans[i]);
  return 0;
}