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;
typedef long long ll;
const int lim = 4e5+5;
int evsize = 0;
vector<int> events[lim];

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

	int n,q;
	cin>>n>>q;
	for(int i=0;i<n;i++)
	{
		int s,t,x;
		cin>>s>>t>>x;
		s -= x;
		t -= x;
		vector<int> add = {s,+1,x};
		vector<int> del = {t,-1,x};
		events[evsize++] = add;
		events[evsize++] = del;
	}
	sort(events,events+evsize);

	set<int> minx;
	int eid = 0;
	for(int i=0;i<q;i++)
	{
		int d;
		cin>>d;
		while(eid<evsize && events[eid][0]<=d)
		{
			if(events[eid][1]==+1)
				minx.insert(events[eid][2]);
			if(events[eid][1]==-1)
				minx.erase(events[eid][2]);
			eid++;
		}

		if(minx.empty())
			cout<<-1<<'\n';
		else
			cout<<*minx.begin()<<'\n';
	}

	return 0;
}