You are given an array a consisting of n distinct elements and an integer k. Each element in the array is a non-negative integer not exceeding 2^k-1.

Let's define the XOR distance for a number x as the value of 

$$$f(x) = min_{i = 1}^{n} min_{j = i + 1}^{n} |(a_i ⊕ x) - (a_j ⊕ x)|,$$$

where ⊕ denotes [the bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

For every integer x from 0 to 2^k-1, you have to calculate f(x).

Input

The first line contains two integers n and k (1 ≤ k ≤ 19; 2 ≤ n ≤ 2^k).

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^k-1). All these integers are distinct.

Output

Print 2^k integers. The i-th of them should be equal to f(i-1).

Examples

Input


3 3
6 0 3


Output


3 1 1 2 2 1 1 3 


Input


3 4
13 4 2


Output


2 2 6 6 3 1 2 2 2 2 1 3 6 6 2 2 

Note

Consider the first example:

  * for x = 0, if we apply bitwise XOR to the elements of the array with x, we get the array [6, 0, 3], and the minimum absolute difference of two elements is 3; 
  * for x = 1, if we apply bitwise XOR to the elements of the array with x, we get the array [7, 1, 2], and the minimum absolute difference of two elements is 1; 
  * for x = 2, if we apply bitwise XOR to the elements of the array with x, we get the array [4, 2, 1], and the minimum absolute difference of two elements is 1; 
  * for x = 3, if we apply bitwise XOR to the elements of the array with x, we get the array [5, 3, 0], and the minimum absolute difference of two elements is 2; 
  * for x = 4, if we apply bitwise XOR to the elements of the array with x, we get the array [2, 4, 7], and the minimum absolute difference of two elements is 2; 
  * for x = 5, if we apply bitwise XOR to the elements of the array with x, we get the array [3, 5, 6], and the minimum absolute difference of two elements is 1; 
  * for x = 6, if we apply bitwise XOR to the elements of the array with x, we get the array [0, 6, 5], and the minimum absolute difference of two elements is 1; 
  * for x = 7, if we apply bitwise XOR to the elements of the array with x, we get the array [1, 7, 4], and the minimum absolute difference of two elements is 3. 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,K,a[530005],c[10600005][2],rt=1,tot=1,mn[21][530005];
void Insert(int p){
	int x=rt,m[24]={0};
	while(p)m[++m[0]]=p%2,p/=2;
	for(int i=K;i>=1;i--){
		if(!c[x][m[i]])c[x][m[i]]=++tot;
		x=c[x][m[i]];
	}
}
int Qmi(int p,int x,int dep){
	if(dep==K)return 0;
	if(c[p][(x>>K-dep-1)&1])return Qmi(c[p][(x>>K-dep-1)&1],x,dep+1);
	if(c[p][((x>>K-dep-1)&1)^1])return Qmi(c[p][((x>>K-dep-1)&1)^1],x,dep+1)+(1<<K-dep-1);
	return assert(0),0;
}
int Qmx(int p,int x,int dep){
	if(dep==K)return 0;
	if(c[p][((x>>K-dep-1)&1)^1])return Qmx(c[p][((x>>K-dep-1)&1)^1],x,dep+1)+(1<<K-dep-1);
	if(c[p][(x>>K-dep-1)&1])return Qmx(c[p][(x>>K-dep-1)&1],x,dep+1);
	return assert(0),0;
}
void dfs(int x,int dep){
	if(!c[x][1]||!c[x][0])return ;
	//cout<<x<<' '<<dep<<' '<<c[x][0]<<' '<<c[x][1]<<'\n';
	for(int i=0;i<(1<<K-dep);i++){
		int w=(i>>K-dep-1)&1;
		mn[dep][i]=min(mn[dep][i],Qmi(c[x][w^1],i,dep+1)-Qmx(c[x][w],i,dep+1)+(1<<K-dep-1));
		//cout<<"MN:"<<dep<<' '<<i<<' '<<mn[dep][i]<<'\n';
	}
	dfs(c[x][0],dep+1),dfs(c[x][1],dep+1);
}
int main(){
	scanf("%d%d",&n,&K),memset(mn,0x3f,sizeof(mn));
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),Insert(a[i]);
	dfs(rt,0);
	for(int i=0;i<(1<<K);i++){
		int ans=1e9;
		for(int j=0;j<K;j++)ans=min(ans,mn[j][i&((1<<K-j)-1)]);
		cout<<ans<<' ';
	}
}
