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 N 1000005
typedef long long LL;
struct mzls
{
	int x1,x,op;
}a[N];
bool operator<(mzls a1,mzls b1)
{
	return a1.x1<b1.x1;
}
pair<int,int>a1[N];
int ans[N],n,m;
map<int,int>mp;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		a[i].x1=x-z,a[i].x=z,a[i].op=0;
		a[i+n].x1=y-z,a[i+n].x=z,a[i+n].op=1;
	}
	n*=2;
	sort(a+1,a+n+1);
	for(int i=1;i<=m;i++)
		scanf("%d",&a1[i].first),a1[i].second=i;
	sort(a1+1,a1+m+1);
	int t1=1;
	for(int i=1;i<=m;i++)
	{
		while(t1<=n&&a[t1].x1<=a1[i].first)
		{
			if(a[t1].op==0)
				mp[a[t1].x]++;
			else
			{
				mp[a[t1].x]--;
				if(mp[a[t1].x]==0)
					mp.erase(a[t1].x);
			}
			t1++;
		}
		if(mp.begin()!=mp.end())
			ans[a1[i].second]=mp.begin()->first;
	}
	for(int i=1;i<=m;i++)
		printf("%d\n",ans[i]==0?-1:ans[i]);
}