#include <bits/stdc++.h>
const int BIT = 45;
int main() {
  long long s, x;
  scanf("%lld%lld", &s, &x);
  bool carry = 0;
  long long ans = 1;
  for (int i = 0; i < BIT; ++i) {
    bool hascarry = (s >> (i + 1) & 1) != (x >> (i + 1) & 1);
    if (hascarry && !carry && (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if (!hascarry && carry && (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if ((carry ^ (s >> i & 1)) != (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if (x >> i & 1) {
      ans <<= 1;
    }
    carry = hascarry;
  }
  if (s == x) {
    ans -= 2;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int BIT = 45;
int main() {
  long long s, x;
  scanf("%lld%lld", &s, &x);
  bool carry = 0;
  long long ans = 1;
  for (int i = 0; i < BIT; ++i) {
    bool hascarry = (s >> (i + 1) & 1) != (x >> (i + 1) & 1);
    if (hascarry && !carry && (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if (!hascarry && carry && (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if ((carry ^ (s >> i & 1)) != (x >> i & 1)) {
      puts("0");
      return 0;
    }
    if (x >> i & 1) {
      ans <<= 1;
    }
    carry = hascarry;
  }
  if (s == x) {
    ans -= 2;
  }
  printf("%lld\n", ans);
  return 0;
}

```