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>
#define rep(i, n) for (ll i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll INF = ll(1e18)+1;
int main(){
    ll n, q;
    cin >> n >> q;
    vector<vector<ll>> xst;
    rep(i, n){
        ll s, t, x;
        cin >> s >> t >> x;
        xst.push_back({x,s,t});
    }
    sort(xst.begin(), xst.end());
    set<P> s;
    rep(i, q){
        ll d;
        cin >> d;
        s.insert(P(d, i));
    }
    vector<ll> ans(q, -1);
    for(auto i:xst){
        int x, l, r;
        x = i[0]; l = i[1]-x; r = i[2]-x;
        auto it = s.lower_bound(P(l, -1));
        while(it != s.end()){
            if(it->first >= r)break;
            else{
                ans[it->second] = x;
                s.erase(it++);
            }
        }
    }
    rep(i, q){
        cout << ans[i] << endl;
    }
    return 0;
}