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,m,n) for(int i=m;i<n;i++)
typedef pair<ll,int> P;

int main(){
  int n,q;
  cin>>n>>q;
  ll s[n],t[n],x[n];
  rep(i,0,n) cin>>s[i]>>t[i]>>x[i];
  ll d[q];
  rep(i,0,q) cin>>d[i];
  
  set<P> st;
  rep(i,0,q){
    st.insert(P(d[i],i));
  }
  
  ll ans[q];
  rep(i,0,q) ans[i]=-1;
  
  vector<ll> a[n];
  rep(i,0,n) a[i]={x[i],s[i],t[i]};
  sort(a,a+n);
  
  rep(i,0,n){
    auto it=st.lower_bound(P(a[i][1]-a[i][0],-1));
    while(it!=st.end()){
      if(it->first>=a[i][2]-a[i][0]) break;
      ans[it->second]=a[i][0];
      st.erase(it++);
    }
  }
  rep(i,0,q) cout<<ans[i]<<endl;
}