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

#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
#define FOR(i, m, n) for(int i = (int)(m); i < (int)(n); ++i)

int main()
{
    int N, Q;
    cin >> N >> Q;
    vector<vector<int>> data(N, vector<int>(3));
    REP(i, N){
        cin >> data[i][1] >> data[i][2] >> data[i][0];
        data[i][1] -= data[i][0];
        data[i][2] -= data[i][0];
    }
    set<pair<int,int>> s;
    REP(i, Q){
        int D;
        cin >> D;
        s.insert({D,i});
    }

    sort(data.begin(), data.end());

    vector<int> ans(Q, -1);
    REP(i, N){
        auto itr = s.lower_bound({data[i][1], -1});
        while(itr != s.end() and itr != s.upper_bound({data[i][2], -1})){
            ans[(*itr).second] = data[i][0];
            s.erase(itr++);
        }
    }

    REP(i, Q) cout << ans[i] << endl;

    return 0;
}