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 <algorithm>
#include <utility>
#include <set>
using namespace std;

pair<int, int> r[600009];
multiset<int> st;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int n, q; cin >> n >> q;
    for (int i = 0; i < n; i++) {
        int s, t, x; cin >> s >> t >> x;
        r[i * 2] = make_pair(s - x, x); r[i * 2 + 1] = make_pair(t - x, -x);
    }
    for (int i = 0; i < q; i++) {
        int d; cin >> d;
        r[n * 2 + i] = make_pair(d, 1000000009);
    }
    sort(r, r + n * 2 + q);
    for (int i = 0; i < n * 2 + q; i++) {
        if (r[i].second == 1000000009) {
            if (st.empty()) cout << -1 << '\n';
            else cout << *(st.begin()) << '\n';
        }
        else if (r[i].second < 0) st.erase(st.find(-r[i].second));
        else 
            st.insert(r[i].second);
    }
    return 0;
}
