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;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;

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

int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);

 ll n,q;
  cin >> n >> q;
  vector<tuple<ll,ll,ll>> v;
  set<ll> st;
  rep(i,0,n) {
   ll s,t,x;
    cin >> s >> t >> x;
    v.push_back(make_tuple(s-x,1,x));
    v.push_back(make_tuple(t-x,-1,x));
  }

   sort(v.begin(),v.end());
  
   ll index=0;
  rep(i,0,q){
   ll d;
    cin >> d;
    
     while(get<0>(v[index])<=d&&index<v.size()){
       if(get<1>(v[index])==1) st.insert(get<2>(v[index]));
       else st.erase(get<2>(v[index]));
       index++; 
     }
      if(!st.empty()) cout << *st.begin() << endl;
      else cout << -1 << endl;
  }

}
