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 <algorithm>
#include <iostream>
#include <set>
using namespace std;
typedef pair<int, int> P;

int main()
{
    int n, q;
    cin >> n >> q;
    int s[200005], t[200005], x[200005];
    P p[400005];
    for(int i = 0; i < n; i++){
        cin >> s[i] >> t[i] >> x[i];
        p[i * 2] = P(s[i] - x[i], x[i]);
        p[i * 2 + 1] = P(t[i] - x[i], -x[i]);
    }
    sort(p, p + n * 2);
    set<int> st;
    int i = 0;
    while(q--){
        int d;
        cin >> d;
        while(i < n * 2 && p[i].first <= d){
            if(p[i].second > 0) st.insert(p[i].second);
            else st.erase(-p[i].second);
            i++;
        }
        if(st.empty()) cout << -1 << endl;
        else cout << *st.begin() << endl;
    }
}
