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 <algorithm>
#include <vector>
#include <set>
using namespace std;
typedef long long int ll;

int ans[200005];
set<pair<ll,int> > S;
int main(){
    int n,q; cin >> n >> q;
    vector<pair<ll,pair<ll,ll> > > v;
    for(int i=0;i<n;i++){
        ll s,t,x; cin >> s >> t >> x;
        v.push_back(make_pair(x,make_pair(s-x,t-x-1)));
    }
    for(int i=0;i<q;i++){
        ll d; cin >> d;
        S.insert(make_pair(d,i));
        ans[i]=-1;
    }
    sort(v.begin(),v.end());
    for(int i=0;i<v.size();i++){
        if(S.empty())continue;
        ll l=v[i].second.first;
        ll r=v[i].second.second;
        auto iter=S.lower_bound(make_pair(l,-1));
        while(S.size()&&iter!=S.end()&&(*iter).first<=r){
            ans[(*iter).second]=v[i].first;
            iter=S.erase(iter);
        }
    }
    for(int i=0;i<q;i++){
        cout << ans[i] << endl;
    }
}