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<cstring>
#include<algorithm>
#include<map>
using namespace std;

struct point{
    int s;
    int t;
    int x;
};

int main(){
    int n, q;
    cin >> n >> q;
    point p[n];
    for(int i=0;i<n;i++) cin >> p[i].s >> p[i].t >> p[i].x;
    sort(p,p+n,[](const point& l, const point& r){
        return l.x < r.x;     
    });

    map<int, int> m;
    for(int i=0;i<q;i++) {
        int d;
        cin >> d;
        m[d] = i;
    }

    int ans[q];
    memset(ans, -1, sizeof(ans));
    for(int i=0;i<n;i++){
        auto itr = m.lower_bound(p[i].s-p[i].x);
        int end = p[i].t-p[i].x;
        while(itr != m.end() && itr->first < end) {
            ans[itr->second]=p[i].x;
            m.erase(itr++);
        }
    }

    for(int i=0;i<q;i++) cout << ans[i] << endl;
    return 0;
}
