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>
#include<iostream>
#include<fstream>
#define ll long long
using namespace std;

typedef tuple<int,int,int> T;

int main(){
    int N, Q;  cin >> N >> Q;
    vector<T> A(N);
    for(int i=0; i<N; i++){
        int S, T, X;  cin >> S >> T >> X;
        A[i] = make_tuple(X, S-X, T-X);
    }
    sort(A.begin(), A.end());

    set<pair<int,int>> D;
    for(int i=0; i<Q; i++){
        int d;  cin >> d;
        D.insert({d,i});
    }

    vector<int> ans(Q,-1);
    for(int i=0; i<N; i++){
        int L, R, X;
        tie(X,L,R) = A[i];
        auto itr = D.lower_bound({L, -1});
        while(itr != D.end()){
            if(itr->first >= R) break;
            ans[itr->second] = X;
            itr = D.erase(itr);
        }
    }

    for(int i=0; i<Q; i++){
        cout << ans[i] << endl;
    }
}
