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;
using ll = long long;
#define rep(i,n) for(int(i)=0;(i)<(n);(i)++)
int main(){
    int n,q; cin>>n>>q;
    vector<tuple<ll,ll,ll>> tap;
    rep(i,n){
        ll s,t,x; cin>>s>>t>>x;
        tap.push_back(make_tuple(s-x,0,x));
        tap.push_back(make_tuple(t-x,-1,x));
    }
    rep(i,q){
        ll d; cin>>d;
        tap.push_back(make_tuple(d,1,i));
    }
    sort(tap.begin(),tap.end());
    multiset<ll> se; 
    vector <ll> ans(q);
    rep(i,tap.size()){
        ll c=get<1>(tap[i]), v=get<2>(tap[i]);
        if(c==1){if(se.empty()) {ans[v]=-1;} else ans[v]=*(se.lower_bound(0));}
        else if(!c) se.insert(v);
        else se.erase(v);
    }
    rep(i,q) cout<<ans[i]<<endl;
}
