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 rep(i, n) for (int i = 0; i < (int)n; i++)

int N, Q, S, T, X, D;
set<pair<int, pair<int, int>>> events; // Ordena los eventos del sweep line
multiset<int> active; // Guarda alturas ordenadas de segmentos activos
vector<int> ans;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> N >> Q;

    rep(i, N)
    {
        cin >> S >> T >> X;
        events.insert({T - X, {0, X}});
        events.insert({S - X, {1, X}});

        if (S - X < 0 && T - X >= 0) active.insert(X);
    }

    rep(i, Q) { cin >> D; events.insert({D, {2, i}}); }

    ans.assign(Q, -1);

    for (auto e : events)
    {
        T = e.second.first;
        X = e.second.second;

        if (T == 0) active.erase(X);
        if (T == 1) active.insert(X);
        if (T == 2 && !active.empty()) ans[X] = *active.begin();
    }

    rep(i, Q) cout << ans[i] << '\n';
}