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 tup = pair<pair<int, int>, int>;

int main() {
    int N, Q;
    cin >> N >> Q;
    vector<tup> E;
    for(int i=0; i<N; ++i) {
        int S, T, X;
        cin >> S >> T >> X;
        E.push_back({{S - X, 1}, X});
        E.push_back({{T - X, -1}, X});
    }
    sort(begin(E), end(E));

    vector<int> D(Q);
    for(int i=0; i<Q; ++i) cin >> D[i];

    set<int> S;

    int cur = 0;
    for(int i=0; i<Q; ++i) {
        while(cur < 2*N && E[cur].first.first <= D[i]) {
            if(E[cur].first.second == 1) S.insert(E[cur].second);
            else S.erase(E[cur].second);
            ++cur;
        }

        if(S.empty()) cout << -1 << endl;
        else cout << *begin(S) << endl;
    }
}