## SPCO - Gopu and Counting Bitwise Prime Numbers

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A positive integer is said to be bitwise prime if the sum of all the bits in its binary representation is a prime number.

You are given two integers a and b. You have to output number of bitwise prime numbers between a and b (inclusive).

### Input

First line contains T : number of test cases. (1 <= T <= 10^5)

For next T lines, each test case contains two space separated integers a and b. (a <= b). 1 <= a, b <= 10^19.

### Output

For each test case, print the number of bitwise prime numbers between a and b (inclusive).

### Example

```Input:
21 21 3

Output:
01 ```

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;

#define MOD (ll)1000000007
#define pb   push_back
#define EPS 1e-9
#define FOR(i,n)  for(int i = 0;i < n; i++)
#define FORE(i,a,b) for(int i = a;i <= b; i++)
#define pi(a)   printf("%d\n", a)
#define all(c)  c.begin(), c.end()
#define tr(container, it)   for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
#define sdi(a, b)   si(a);si(b)
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl '\n'

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}

const int MAXN = 65;
int isPrime[MAXN];	//isPrime[i] = 0 indicates i is prime.
void sieve(){
//if even check itself while calling. This function will only tells wether a number is prime or not(not for even numbers).
isPrime[0] = isPrime[1] = 1;
for(int i = 3; i*i <= MAXN; i+=2){
if(isPrime[i] == 0){
if(i*(ll)1*i <= MAXN){
for(int j = i*i; j <= MAXN; j += (2*i)){
isPrime[j] = 1;
}
}
}
}
}

int main(){
io;
sieve();
int t;
cin >> t;
while(t--){
double a, b;
cin >> a >> b;
int count = 0;
int bitsA = ((double)log2(a) + 1);
int bitsB = ((double)log2(b) + 1);
for(int i = bitsA; i < bitsB; i++){
if(!isPrime[i]){
count++;
}
}
//if b+1 is power of two and bitsB is prime then count+1
ll rem = b+1;
if(!(rem&(rem-1)) && !isPrime[bitsB]>{
count++;
}
cout << count << endl;
}
return 0;
}``````
Copy The Code &

Input

cmd
2
1 2
1 3

Output

cmd
0
1