# Prime XOR

### Problem Statement :

```Penny has an array of n integers, [a0,a1, a2, ...., an-1]. She wants to find the number of unique multisets she can form using elements from the array such that the bitwise XOR of all the elements of the multiset is a prime number. Recall that a multiset is a set which can contain duplicate elements.

Given q queries where each query consists of an array of integers, can you help Penny find and print the number of valid multisets for each array? As these values can be quite large, modulo each answer by  10^9 + 7 before printing it on a new line.

Input Format

The first line contains a single integer, q, denoting the number of queries. The 2.q subsequent lines describe each query in the following format:

1.The first line contains a single integer, n, denoting the number of integers in the array.
2.The second line contains n space-separated integers describing the respective values of a0, a1,...,an-1.
Constraints

1 <= q <= 10
1 <= n <= 100000
3500 <= ai <= 4500
Output Format

On a new line for each query, print a single integer denoting the number of unique multisets Penny can construct using numbers from the array such that the bitwise XOR of all the multiset's elements is prime. As this value is quite large, your answer must be modulo 10^9 + 7.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
typedef long long ll;
ll n,i,j,k,m,x;
ll primes, b, dp;
set<ll> f;
set<ll>::iterator itr;
int main()
{
//freopen("input.txt","r",stdin);
for (i = 2; i <= 10000; i++)
primes[i] = 1;
for (i = 2; i <= 10000; i++)
if (primes[i])
for (j = i*2; j <= 10000; j+=i)
primes[j] = 0;
ll q;
cin >> q;
while (q--)
{
cin >> n;
for (i = 3500; i <= 4500; i++)
b[i] = 0;
for (i = 0; i < n; i++)
{
scanf("%lld",&x);
b[x]++;
}
for (i = 0; i <= 1001; i++)
for (j = 0; j <= 8192; j++)
dp[i][j] = 0;
dp = 1;
for (i = 0; i <= 1000; i++)
for (j = 0; j < 8192; j++)
{
dp[i+1][j^(i+3500)] = (dp[i+1][j^(i+3500)] + dp[i][j]*((b[i+3500]+1)/2))%MOD;
dp[i+1][j] = (dp[i+1][j] + dp[i][j]*(b[i+3500]/2+1))%MOD;
}
ll ans = 0;
for (j = 0; j < 8192; j++)
if (primes[j])
ans += dp[j];
cout << ans%MOD << endl;
}
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static int mod = 1000000007;
public static boolean[] prime = new boolean[ 10000 ];
public static long[] dp = new long[ 1 << 13 ];
public static long[] dpp = new long[ 1 << 13 ];
public static long[] val = new long[ 5000 ];
public static ArrayList< Integer > list = new ArrayList< Integer >();;
public static void makeArray()
{
for( int i = 0; i < 10000; i++ )
prime[ i ] = true;
prime[ 0 ] = prime[ 1 ] = false;
for( int i = 2; i < 10000; i++ )
{
if( !prime[ i ] )
continue;
for( int j = 2 * i; j < 10000; j += i)
prime[ j ] = false;
}
}

public static void check()
{
for( int i = 0; i < ( 1 << 13 ); i++ )
{
dp[ i ] = dpp[ i ] = 0;
if( i < 4999 )
val[ i ] = 0;
}
list.clear();
}
public static void main(String[] args) {

Scanner input = new Scanner( System.in );
makeArray();
int q = input.nextInt();
while( q > 0 )
{
check();
int n = input.nextInt();
long ans = 0l;
for( int i = 0;  i < n; i++ )
{
int x = input.nextInt();
if( val[ x ] == 0 )
val[ x ]++;
}

n = list.size();

for( int i = 0; i < n; i++ )
{
for( int j = 0; j < ( 1 << 13 ); j++ )
{
long free = val[ list.get( i ) ];
long odd = free / 2;

odd = ( free % 2 == 0 ) ? odd : odd + 1;
long even = 1 + free / 2;

if( i != 0 )
dp[ j ] = dpp[ j ] * even + dpp[ j ^ list.get( i ) ] * odd;
else
{
dp[ list.get( i ) ] = odd;
dp[ 0 ] = even;
break;
}
if( dp[ j ] >= mod )
dp[ j ] %= mod;
}

for( int j = 0; j < ( 1 << 13 ); j++ )
{
if( ( i == n - 1 ) && prime[ j ] )
{
ans += dp[ j ];
if( ans >= mod )
ans %= mod;
}
dpp[ j ] = dp[ j ];
dp[ j ] = 0;
}
}
System.out.println( ans );
q--;
}
}
}

In C :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MOD 1000000007
void gen_primes(int max,int*primes);
int a,p;
long long dp;

int main(){
int q,n,x,i,j;
long long ans;
gen_primes(8191,p);
scanf("%d",&q);
while(q--){
memset(a,0,sizeof(a));
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d",&x);
a[x-3500]++;
}
for(i=0;i<8192;i++)
dp[i]=0;
dp=1;
for(i=0;i<1001;i++){
for(j=0;j<8192;j++)
dp[i+1][j]=dp[i][j];
if(a[i])
for(j=0;j<8192;j++){
dp[i+1][j^(i+3500)]=(dp[i+1][j^(i+3500)]+dp[i][j]*((a[i]+1)/2))%MOD;
dp[i+1][j]=(dp[i+1][j]+dp[i][j]*(a[i]/2))%MOD;
}
}
for(i=ans=0;i<8192;i++)
if(p[i])
ans=(ans+dp[i])%MOD;
printf("%lld\n",ans);
}
return 0;
}
void gen_primes(int max,int*primes){
int i,j;
for(i=0;i<=max;++i)
primes[i]=1;
primes=primes=0;
for(i=2;i*i<=max;++i){
if(!primes[i])
continue;
for(j=2;i*j<=max;++j)
primes[i*j]=0;
}
}

In Python3 :

import math
import os
import random
import re
import sys
import itertools
from itertools import combinations
from collections import Counter

def isPrime(x):
if x == 1:
return False

if x == 2:
return True

for d in range(2, max(2, int(x**0.5)) + 1):
if x%d == 0:
return False
return True

def primeXor(a):

count=0
c=Counter(a)
dp=*8192
dp=1
cache=[]
for e in c.keys():
even=c[e]//2+1
odd=(c[e]+1)//2
dp=[(dp[i]*even +dp[i^e]*odd)%(10**9+7) for i in range(8192)]

for j in range(8192):
if isPrime(j):
count+=dp[j]
if count>(10**9+7):
count=count%(10**9+7)

return count

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

q = int(input())

for q_itr in range(q):
n = int(input())

a = list(map(int, input().rstrip().split()))

result = primeXor(a)

fptr.write(str(result) + '\n')

fptr.close()```
```

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