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 <vector>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <math.h>
#include<iomanip>
using namespace std;

int N, Q;

int main() {
    cin >> N >> Q;
    vector<pair<pair<long long, bool>, long long>> event;
    long long s, t, x;
    for (int i = 0; i < N; i++) {
        cin >> s >> t >> x;
        event.push_back({{s - x, true}, x});
        event.push_back({{t - x, false}, x});
    }
    long long d;
    for (int i = 0; i < Q; i++) {
        cin >> d;
        event.push_back({{d, true}, 1e9 + 1});
    }
    sort(event.begin(), event.end());
    set<long long> S;
    for (auto e : event) {
        x = e.second;
        if (x > 1e9) {
            cout << (S.empty()? -1 : *S.begin()) << endl;
            continue;
        }
        if (e.first.second) {
            S.insert(x);
        } else {
            S.erase(x);
        }
    }
}
