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;
#define ll long long

int main(){
    int N, Q;
    cin >> N >> Q;
    vector<pair<int, pair<int, int>>> event;
    for (int i = 0; i < N; i++) {
        int s, t, x;
        scanf("%d", &s);
        scanf("%d", &t);
        scanf("%d", &x);
        event.push_back({s-x, {1, x}});
        event.push_back({t-x, {-1, x}});
    }
    sort(event.begin(), event.end());
    set<int> stop;
    auto it = event.begin();
    for (int i = 0; i < Q; i++) {
        int d;
        scanf("%d", &d);
        while (it != event.end() && (*it).first <= d) {
            int p = (*it).second.first;
            int q = (*it).second.second;
            if (p == 1) stop.insert(q);
            else stop.erase(q);
            it++;
        }
        if (!stop.empty()) printf("%d\n", *stop.begin());
        else printf("-1\n");
    }
    return 0;
}
