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 ll long long
using namespace std;

int main() {
    ll n,q;
    cin>>n>>q;
    vector<tuple<ll,ll,ll>> event;
    for(ll i=0;i<n;i++) {
        ll s,t,x;
        cin >>s>>t>>x;
        event.push_back(make_tuple(s-x,1,x));
        event.push_back(make_tuple(t-x,-1,x));
    }
    sort(event.begin(),event.end());
    set<ll> stop;
    ll i=0;
    for(ll j=0;j<q;j++) {
        ll d;
        cin>>d;
        while(i<2*n && get<0>(event[i])<=d) {
            if(get<1>(event[i])==1) {
                stop.insert(get<2>(event[i]));
            }else {
                stop.erase(get<2>(event[i]));
            }
            i++;
        }
        if(stop.empty()) {
            cout << -1 << endl;
        }else {
            cout << (ll)*stop.begin() << endl;
        }
    }
}
