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;

int main(int argc, char const *argv[])
{
	int n,q;
	cin>>n>>q;
	vector<pair<pair<long long,int>,long long>> v;
	long long s,t,x;
	for (int i = 0; i < n; ++i)
	{
		cin>>s>>t>>x;
		v.emplace_back(make_pair(make_pair(s-x,-1),x));
		v.emplace_back(make_pair(make_pair(t-x,0),x));
	}
	long long d;
	for (int i = 0; i < q; ++i)
	{
		cin>>d;
		v.emplace_back(make_pair(make_pair(d,1),i));
	}
	sort(v.begin(), v.end());
	multiset<long long> memo={};
	for (auto i = v.begin(); i != v.end(); ++i)
	{
		auto p=*i;
		if (p.first.second==-1)
		{
			memo.insert(p.second);
		}
		else if (p.first.second==0)
		{
			memo.erase(memo.lower_bound(p.second));
		}
		else
		{
			if (memo.size()>0)
			{
				cout<<*memo.begin()<<endl;
			}
			else
			{
				cout<<"-1\n";
			}
		}
	}
	return 0;
}
