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>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i < n; i++)
#define rep3(i, x, n) for(int i = x; i > n; i--)
long long MOD = 1e9+7;
long long INF = 1e15;
typedef pair<long long, long long> P;
typedef long long ll;

int main(){
    long long N, Q;
    cin >> N >> Q;
    vector<P> vec;
    rep(i, N){
        ll S, T, X;
        cin >> S >> T >> X;
        vec.push_back(P(S-X, -X));
        vec.push_back(P(T-1-X, X));
    }
    rep(i, Q){
        ll D;
        cin >> D;
        vec.push_back(P(D, 0));
    }
    sort(vec.begin(), vec.end());
    set<ll> s;
    rep(i, vec.size()){
        P p = vec[i];
        if(p.second < 0) s.insert(-p.second);
        else if(p.second > 0) s.erase(p.second);
        else{
            if(s.size() == 0) cout << -1 << endl;
            else cout << *s.begin() << endl;
        }
    }
}