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)
typedef long long ll ;
typedef tuple<int, int, int> repair_info;
typedef pair<int,int> P;

int main(){
  int N, Q;
  cin >> N >> Q;
  vector<repair_info> a;
  for (auto i = 0; i < N; i++)
  {
    ll s, t, x;
    cin >> s >> t >> x;
    if(s-x<0)s = 0;
    else s = s-x;
    if(t-x<0)t = 0;
    else t = t-x;
    a.push_back(repair_info(x,s,t));
  }
  sort(a.begin(), a.end());
  int d[Q];
  rep(i,Q)cin >> d[i];
  set<P> S;
  rep(i,Q)S.insert(P(d[i],i));
  vector <int> ans(Q,-1);
  rep(i,N){
    int x,l,r;
    tie(x,l,r)= a[i];
    auto it = S.lower_bound(P(l,-1));
    while(it != S.end()){
      if(it->first>=r)break;
      ans[it->second] = x;
      S.erase(it++);
    }
  }
  rep(i,Q)cout << ans[i] << endl;
  // rep(i,N){
  //   int& t = get<0>(a[i]);
  //   std::cout << t << std::endl;
  // }
}
