#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int maxn = 1e5 + 9;
int main() {
  long long s, x;
  while (cin >> s >> x) {
    bool f = 0;
    if (x == s) f = 1;
    if (s < x) {
      puts("0");
      return 0;
    }
    s -= x;
    if (s % 2) {
      puts("0");
      return 0;
    }
    s /= 2;
    long long ans = 1;
    for (int i = 0; i < 61; i++) {
      int x1 = (s >> i) & 1;
      int x2 = (x >> i) & 1;
      if (x1 == x2 && x1 == 1) {
        puts("0");
        return 0;
      }
      if (x2 == 1) ans *= 2;
    }
    if (!f)
      cout << ans << endl;
    else
      cout << ans - 2 << endl;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int mod = 998244353;
const int maxn = 1e5 + 9;
int main() {
  long long s, x;
  while (cin >> s >> x) {
    bool f = 0;
    if (x == s) f = 1;
    if (s < x) {
      puts("0");
      return 0;
    }
    s -= x;
    if (s % 2) {
      puts("0");
      return 0;
    }
    s /= 2;
    long long ans = 1;
    for (int i = 0; i < 61; i++) {
      int x1 = (s >> i) & 1;
      int x2 = (x >> i) & 1;
      if (x1 == x2 && x1 == 1) {
        puts("0");
        return 0;
      }
      if (x2 == 1) ans *= 2;
    }
    if (!f)
      cout << ans << endl;
    else
      cout << ans - 2 << endl;
  }
  return 0;
}

```