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



int main()
{
    int N, Q;
    cin >> N >> Q;
    vector<vector<int>> P(N, vector<int>(3));
    set<pair<int,int>> D;
    for( int i = 0; i < N; ++i ){
        cin >> P[i][1] >> P[i][2] >> P[i][0];
        P[i][1] -= P[i][0];
        P[i][2] -= P[i][0];
    }
    for( int i = 0; i < Q; ++i ){
        int d;
        cin >> d;
        D.insert( make_pair(d, i) );
    }


    // X の小さい順にソート
    sort(P.begin(), P.end());

    vector<int> ans(Q);
    for( int i = 0; i < N; ++i ){
        auto itr = D.lower_bound( make_pair(P[i][1],-1) );
        while( itr->first < P[i][2] and itr != D.end() ){
            ans[itr->second] = P[i][0];
            D.erase(itr++);
        }
    }
    for(auto v: D) ans[v.second] = -1;

    for(auto v: ans) cout << v << endl;
    
    return 0;
}