#include<bits/stdc++.h>
using namespace std;

int main(){
    typedef pair<int, int> P;

    int n, q;
    cin >> n >> q;

    vector<vector<int>> a(n, vector<int>(3));
    int s, t, x;
    for (int i = 0; i < n; i++){
        cin >> s >> t >> x;
        a[i][0] = x;
        a[i][1] = s - x;
        a[i][2] = t - x;
    }
    sort(a.begin(), a.end());

    set<P> p;
    int d;
    for (int i = 0; i < q; i++){
        cin >> d;
        p.insert({d, i});
    }

    vector<int> ans(q, - 1);
    for (int i = 0; i < n; i++){
        x = a[i][0];
        s = a[i][1];
        t = a[i][2];
        auto it = p.lower_bound(P(s, -1));
        while (it != p.end()){
            if (it -> first >= t) break;
            ans[it -> second] = x;
            p.erase(it++);
        }
    }

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

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int main(){
    typedef pair<int, int> P;

    int n, q;
    cin >> n >> q;

    vector<vector<int>> a(n, vector<int>(3));
    int s, t, x;
    for (int i = 0; i < n; i++){
        cin >> s >> t >> x;
        a[i][0] = x;
        a[i][1] = s - x;
        a[i][2] = t - x;
    }
    sort(a.begin(), a.end());

    set<P> p;
    int d;
    for (int i = 0; i < q; i++){
        cin >> d;
        p.insert({d, i});
    }

    vector<int> ans(q, - 1);
    for (int i = 0; i < n; i++){
        x = a[i][0];
        s = a[i][1];
        t = a[i][2];
        auto it = p.lower_bound(P(s, -1));
        while (it != p.end()){
            if (it -> first >= t) break;
            ans[it -> second] = x;
            p.erase(it++);
        }
    }

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