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 (long long i = 0; i < (long long)(n); i++)
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
using vi=vector<ll>;
using vvi=vector<vector<ll>>;
int main(){
  int n,q;cin>>n>>q;
  vector<PP> p(n);
  rep(i,n){
    int s,t,x;scanf("%d %d %d",&s,&t,&x);
    get<2>(p[i])=x-t+1;
    get<1>(p[i])=x-s;
    get<0>(p[i])=x;
  }
  sort(p.begin(),p.end());
  set<P> s;
  vi ans(q,-1);
  rep(i,q){
    int a;scanf("%d",&a);
    s.insert(P(-a,i));
  }
  rep(i,n){
    if(s.size()==0)break;
    int a=get<2>(p[i]),b=get<1>(p[i]);
    auto from=s.lower_bound(P(a,-1));
    while(from!=s.end()){
      if((from->first)>get<1>(p[i]))break;
      ans[from->second]=get<0>(p[i]);
      s.erase(from++);
    }
  }
  rep(i,q)printf("%d\n",ans[i]);
}
