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;
typedef tuple<int, int, int> T;
int main() {
    int N, Q;
    cin >> N >> Q;
    vector<T> v(N);
    for (int i = 0; i < N; i++) {
        int s, t, x;
        cin >> s >> t >> x;
        v[i] = T(x, s, t);
    }
    sort(v.begin(), v.end());
    vector<int> D(Q);
    set<int> se;
    se.insert(2e9);
    for (int i = 0; i < Q; i++) {
        cin >> D[i];
        se.insert(D[i]);
    }
    map<int, int> ma;
    for (int i = 0; i < N; i++) {
        int x = get<0>(v[i]), s = get<1>(v[i]), t = get<2>(v[i]);
        int l = s - x;
        int r = t - x;
        while (1) {
            auto itr = se.lower_bound(l);
            if (*itr >= r) break;
            ma[*itr] = x;
            se.erase(*itr);
        }
    }
    for (int i = 0; i < Q; i++) {
        int p = ma[D[i]];
        cout << (p ? p : -1) << endl;
    }
}
