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

typedef pair <int, int> ii;

int n, q;
vector <ii> seq;
set <int> S;

int main()
{
    scanf("%d %d", &n, &q);
    for (int i = 0; i < n; i++) {
        int s, t, x; scanf("%d %d %d", &s, &t, &x);
        seq.push_back(ii(s - x, x));
        seq.push_back(ii(t - x, -x));
    }
    sort(seq.begin(), seq.end());
    int pnt = 0;
    while (q--) {
        int d; scanf("%d", &d);
        while (pnt < seq.size() && seq[pnt].first <= d) {
            if (seq[pnt].second < 0) S.erase(-seq[pnt].second);
            else S.insert(seq[pnt].second);
            pnt++;
        }
        if (S.empty()) printf("-1\n");
        else printf("%d\n", *S.begin());
    }
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

typedef pair <int, int> ii;

int n, q;
vector <ii> seq;
set <int> S;

int main()
{
    scanf("%d %d", &n, &q);
    for (int i = 0; i < n; i++) {
        int s, t, x; scanf("%d %d %d", &s, &t, &x);
        seq.push_back(ii(s - x, x));
        seq.push_back(ii(t - x, -x));
    }
    sort(seq.begin(), seq.end());
    int pnt = 0;
    while (q--) {
        int d; scanf("%d", &d);
        while (pnt < seq.size() && seq[pnt].first <= d) {
            if (seq[pnt].second < 0) S.erase(-seq[pnt].second);
            else S.insert(seq[pnt].second);
            pnt++;
        }
        if (S.empty()) printf("-1\n");
        else printf("%d\n", *S.begin());
    }
    return 0;
}

```