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;
using ll = long long;
using pi = pair<int, int>;
set<pi> st;
int n,q;
multiset<int> ans;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> q;
    for(int i=1;i<=n;i++)
    {
        int s,t,x;
        cin >> s >> t >> x;
        st.insert({s-x, -x});
        st.insert({t-x, x});
    }
    auto it = st.begin();
    for(int i=1;i<=q;i++){
        int d;
        cin >> d;
        while(it != st.end() && d >= it->first)
        {
            if(it->second > 0)
            {
                auto it2 = ans.find(it->second);
                ans.erase(it2);
            }
            else
                ans.insert(-(it->second));
            it++;
        }
        if(ans.empty()) cout << "-1\n";
        else cout << *ans.begin() << '\n';
    }
}