In this Problem there is one point to observe. I will explain that with the help of example.
Assume there are tw numbers a and b. we are given a=5 and OR(a ,b) =9.
5=0101
OR(a,b) = 9 =1101
Here b’s binary form will be mnop. Here o can not be 1 as in OR(a,b) these bits are equal to zero.
And ‘m’ has to be 1 since in OR(a,b) this bit is 1 but in ‘a’ this bit is 0.
So there is only one case left when in OR(a,b) and a, bit is 1. This case represents n and p in above example. Here n and p can be zero or one. So by Multiplication Rule of Probability there are total 4 possibilities of b.
If we need to find the number of possible values of b then we have to find the number of set bits which are at same position in a and OR(a,b).
This can be done by counting set bits in (a & OR(a,b)).
At last you need to check if given array is sorted or not. If array is sorted then There is no possibility of sequence.
Here you need to check one more condition that b[i]&b[i+1])!=b[i].
My code is here

#include < iostream>

#include < algorithm>

#include < string>

#include < cmath>

#include < cstdlib>

#include < set>

#include < vector>

#include < sstream>

#include < queue>

#include < iomanip>

constexpr auto INF = 9223372036854775807;

typedef long long int ll;

typedef unsigned long long int ull;

typedef unsigned long int ul;

using namespace std;

vector a;

int main()

{

ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

ull test;

cin >> test;

for (ull i = 0; i < test; i += 1)

{

ull n = 0;

cin >> n;

ull ans = 1;

vector<ull> arr(n, 0);

for (ull j = 0; j < n; j += 1)

{

cin >> arr[j];

if ((j>0)&&arr[j]>=arr[j1])

{

// Count set bit in arr[j]&arr[j1]

ull ad = arr[j] & arr[j  1];

ull count = 0;

while (ad)

{

count += ad & 1;

ad >>= 1;

}

ans *= (ull)(pow(2, count));

ans %= 1000000007;

}

if ((j > 0)&&(arr[j] <arr[j1]) )

{

ans = 0;

}

if ((j>0) && ((arr[j] & arr[j  1]) != arr[j  1]))

{

ans = 0;

> }

}

cout << ans << "\n";

}

return 0;

}