#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const double eps = 1e-15;
double PI = 3.14159265359;
const int INF = 1e9;
const int N = 1000001;
;
int n, t;
long long s, x, a, r = 1;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> x;
  if ((s - x) & 1) {
    cout << 0;
    return 0;
  }
  a = (s - x) >> 1;
  if (a & x) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i < 50; i++)
    if (x & 1LL << i) r <<= 1;
  if (!a) r -= 2;
  cout << r;
}

### Prompt

Your task is to create a cpp solution to the following problem:

Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?

Input

The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.

Output

Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.

Examples

Input

9 5


Output

4


Input

3 3


Output

2


Input

5 2


Output

0

Note

In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).

In the second sample, the only solutions are (1, 2) and (2, 1).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const double eps = 1e-15;
double PI = 3.14159265359;
const int INF = 1e9;
const int N = 1000001;
;
int n, t;
long long s, x, a, r = 1;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> x;
  if ((s - x) & 1) {
    cout << 0;
    return 0;
  }
  a = (s - x) >> 1;
  if (a & x) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i < 50; i++)
    if (x & 1LL << i) r <<= 1;
  if (!a) r -= 2;
  cout << r;
}

```