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;
const int N=200005;
vector<pair<int,pair<int,int>>>a;
multiset<int>s;
int main()
{
	int n,q;
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)
	{
		int s,t,x;
		scanf("%d%d%d",&s,&t,&x);
		a.push_back(make_pair(s-x,make_pair(1,x)));
		a.push_back(make_pair(t-x,make_pair(-1,x)));
	}
	sort(a.begin(),a.end());
	int i=0;
	while(q--)
	{
		int d;
		scanf("%d",&d);
		while(a[i].first<=d&&i<a.size())
		{
			if(a[i].second.first==1)
				s.insert(a[i].second.second);
			else
				s.erase(a[i].second.second);
			i++;
		} 
		if(s.size())
			printf("%d\n",*s.begin());
		else
			printf("-1\n");
	}
	return 0;
}
