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 main(){
    int N, Q;
    cin >> N >> Q;
    static vector<int> STX[200000];
    for(int i=0; i<N; i++){
        int s, t, x;
        cin >> s >> t >> x;
        STX[i] = {s-x, t-x, x};
    }
    sort(STX, STX+N);

    typedef vector<int> V;
    int pt = 0;
    // XST
    priority_queue<V, vector<V>, greater<V>> que;
    while(Q--){
        int d;
        cin >> d;
        while(pt < N && STX[pt][0] <= d){
            auto v = STX[pt];
            que.push({v[2], v[0], v[1]});
            pt++;
        }
        while(que.size() && que.top()[2] <= d) que.pop();
        if(que.size()){
            cout << que.top()[0] << endl;
        }else{
            cout << -1 << endl;
        }
    }
    return 0;
}
