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;
using P = pair<int, int>;

int main() {
    int n, q;
    cin >> n >> q;
    vector<pair<P, int>> v;
    for (int i = 0; i < n; ++i) {
        int s, t, x;
        cin >> s >> t >> x;
        v.push_back({ { s - x, 1 }, x });
        v.push_back({ { t - x, -1 }, x });
    }
    sort(v.begin(), v.end());
    set<int> se;
    int j = 0;
    for (int i = 0; i < q; ++i) {
        int d;
        cin >> d;
        while (j < v.size() && v[j].first.first <= d) {
            if (v[j].first.second == 1) {
                se.insert(v[j].second);
            }
            else {
                se.erase(v[j].second);
            }
            ++j;
        }
        if (se.size())
            cout << *se.begin() << '\n';
        else
            cout << "-1\n";
    }

    return 0;
}
