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;
int in(){
    int x;
    cin>>x;
    return x;
}
int main(){
    int n=in(),q=in();
    vector<int>ans(q,-1);
    tuple<int,int,int>work[n];
    for(int i=0;i<n;i++){
        int s=in(),t=in(),x=in();
        work[i]={x,s,t};
    }
    sort(work,work+n);
    set<pair<int,int>>people;
    for(int i=0;i<q;i++)people.insert({in(),i});
    for(int i=0;i<n;i++){
        int x=get<0>(work[i]),s=get<1>(work[i]),t=get<2>(work[i]);
        for(auto itr=people.lower_bound({s-x,0});
        itr!=people.lower_bound({t-x,0});itr=people.erase(itr)){
            ans[(*itr).second]=x;
        }
    }
    for(int i:ans)cout<<i<<endl;
    return 0;
}