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


int main()
{
ll n , q ;
cin >> n >> q ;
vector<pair<ll,ll>> v;
rep(i,n){
  ll s , t , x ;
  cin >> s >> t >> x ;
  s-=x;if(s<0)s=0;
  t-=x;if(t<0)t=0;
  if(t>0){
  v.push_back({s,x});
  v.push_back({t,-x});
  }
}
sort(v.begin(),v.end());
vector<ll> ans(q);
multiset<ll> s;
ll itr=0;
rep(i,q){
  ll d ;
  cin >> d ;
  while(itr<v.size() && d>=v[itr].first){
    if(v[itr].second>0)s.insert(v[itr].second);
    else s.erase(s.find(-1*v[itr].second));
    itr++;
  }
  if(s.size()>0)ans[i]=*s.begin();
  else ans[i]=-1;
}

for(auto an:ans)cout << an << endl;
return 0;
}