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;
#define rep(i,N) for(int (i)=0;(i) < (N); (i)++)
#define all(V) V.begin(), V.end()
using i64 = int_fast64_t;
using P = pair<int,int>;

int main(){
    int N, Q;
    cin >> N >> Q;
    priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> que;

    int s, t, x;
    rep(i,N){
        cin >> s >> t >> x;
        que.push({s-x, x, 1});
        que.push({t-x, x , 0});
    }

    int ans = -1, D;
    set<int> Xs;
    rep(i, Q){
        cin >> D;
        while(!que.empty() && que.top()[0] <= D){
            int t = que.top()[0];
            int x = que.top()[1];
            int is_push = que.top()[2];
            que.pop();

            if(is_push) Xs.insert(x);
            else Xs.erase(x);
        }
        if(Xs.size() == 0) ans = -1;
        else ans = *Xs.begin();

        cout << ans << endl;
    }
}
