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<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
constexpr int mo = 1e9+7;
constexpr int  mod = mo;
constexpr ll inf = 1ll<<62;

int main(){
    int n,q;
    cin>>n>>q;
    vector<tuple<int,int,int>> tar;
    set<pair<int,int>> d;
    for(int i=0;i<n;++i){
        int s,t,x;
        cin>>s>>t>>x;
        tar.emplace_back(x,s,t);
    }
    sort(tar.begin(),tar.end());
    for(int i=0;i<q;++i){
        int di;cin>>di;
        d.emplace(di,i);
    }
    vector<ll> result(q,-1);
    for(int i=0;i<n;++i){
        auto fir = d.lower_bound(make_pair(get<1>(tar[i]) - get<0>(tar[i]),0));
        auto las = d.lower_bound(make_pair(get<2>(tar[i]) - get<0>(tar[i])-1,q));
        for(auto it = fir;it!=las;++it){
                result[it->second] = get<0>(tar[i]);
        }
        d.erase(fir,las);
    }
    for(auto r : result)cout << r << endl;
    return 0;
}

