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;
#define mp make_pair
typedef long long ll;
typedef pair<int,int>pii;
const int N = 2e5;
int ans[N+100];
set<pii>st;
map<int,vector<pii> >m;
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);
		if(s==t)continue;
		m[x].push_back(mp(s,t-1));
	}
	for(int i=1;i<=q;i++){
		int d;
		scanf("%d",&d);
		st.insert(mp(d,i));
	}
	for(auto i:m)
	{
		for(auto j:m[i.first])
		{
			int l = j.first-i.first,r = j.second-i.first;
			auto it = st.lower_bound(mp(l,-1));
			while(it!=st.end()&&it->first<=r)
			{
				ans[it->second] = i.first;
				pii tmp = *it;
				it++;
				st.erase(tmp);
			}
		}
	}
	for(auto it:st)
	{
		ans[it.second] = -1;
	}
	for(int i=1;i<=q;i++){
		printf("%d\n",ans[i]);
	}
	printf("\n");
}
