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>
#define rep(i,n) for(int i=0;i<(n);++i)
using namespace std;
using ll = long long;
int main(){
    int n,q;
    cin >> n >> q;
    vector<array<ll,3>> events;
    rep(i,n){
        ll s,t,x;
        cin >> s >> t >> x;
        events.push_back({s-x,1,x});
        events.push_back({t-x,-1,x});
    }
    sort(events.begin(),events.end(),[](array<ll,3> a,array<ll,3> b){return a[0]>b[0] || (a[0]==b[0] && a[1]>b[1]);});
    set<ll> stop_coord;
    rep(i,q){
        ll d;
        cin >> d;
        while(!events.empty() && (*events.rbegin())[0]<=d){
            if( (*events.rbegin())[1] == 1 )stop_coord.emplace((*events.rbegin())[2]);
            else stop_coord.erase((*events.rbegin())[2]);
            events.pop_back();
        }
        if(stop_coord.empty())cout << "-1" << endl;
        else cout << *stop_coord.begin() << endl;
    }
    return 0;
}