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 <iostream>
#include<bits/stdc++.h>
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,P> PP;
set<ll> se;
ll n,q,a,b,c;
vector<PP> v;
int main(void){
    cin>>n>>q;
    for(int i=0;i<n;i++){
        cin>>a>>b>>c;
        v.push_back({a-c,{0,c}});
        v.push_back({b-c-1,{2,c}});
    }
    while(q--){
        cin>>a;
        v.push_back({a,{1,0}});
    }
    sort(v.begin(),v.end());
    for(auto x:v){
        if(x.S.F==0)se.insert(x.S.S);
        else if(x.S.F==2)se.erase(x.S.S);
        else{
            if(se.size()==0)cout<<-1<<endl;
            else cout<<*se.begin()<<endl;
        }
    }
}
