Problem- Xor it :

Chef has given you a sequence A[1], A[2], ..., A[N] composed of N nonnegative integer numbers. Then, for each pair (i; j) such that 1 <= i < j <= N, we have written a number that equals to A[i] xor A[j] (xor is exclusive or, "xor" in Pascal, "^" in C++). Thus, we have obtained N*(N-1)/2 numbers. Your task is to find K minimal numbers among them. 

Input

The first line of the input contains two space separated integers N and K. Each of the next N lines contains one integer, ith line contains number A[i]. 

Output

In the only line of output print space separated sequence of K numbers, the answer to the problem. Numbers should be in non-decreasing order. 

Constraints

2 ≤ N ≤ 100000 

1 ≤ K ≤ min{250000, N*(N-1)/2} 

0 ≤ A[i] < 231 

Example
Input:
4 5
1 1 3 4

Output:
0 2 2 5 5

Explanation

In the sample input we have 4 numbers: 1, 1, 3, 4. Therefore, there are (4*3)/2 = 6 pairwise XOR's. These XOR's are:
1 xor 1 = 0 (A[1] xor A[2])
1 xor 3 = 2 (A[1] xor A[3])
1 xor 4 = 5 (A[1] xor A[4])
1 xor 3 = 2 (A[2] xor A[3])
1 xor 4 = 5 (A[2] xor A[4])
3 xor 4 = 7 (A[3] xor A[4])
If we sort these numbers we will obtain: 0, 2, 2, 5, 5, 7. The first 5 minimal numbers are: 0, 2, 2, 5, 5.

my solution :

#include<stdio.h>
#include<conio.h>
int main(){
int a[100],b[100],i,j,calc_pwxor,z;
unsigned long int n,k;
printf("input number of terms and minimal number...\n");
scanf("%ld%ld",&n,&k);
printf("input %d numbers.",n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);

calc_pwxor=(n*(n-1))/2;
z=0;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
b[z++]=a[i]^a[j];//printf("%d\t",a[i]^a[j]);
printf("First five minimal number are:");
for(i=0;i<k;i++)
printf("\n\t\t\t%d",b[i]);

return 0;
}