#include <bits/stdc++.h>
long long s, x;
int main() {
  scanf("%I64d%I64d", &s, &x);
  long long cur = 0;
  for (long long i = 0; i < 63; ++i)
    if (x >> i & 1) cur |= 1LL << i;
  cur = s - cur;
  if (cur % 2 != 0 || ((cur / 2) & x) != 0) {
    puts("0");
    return 0;
  }
  cur /= 2;
  int k = 0;
  for (int i = 0; i < 64; ++i)
    if (x >> i & 1) ++k;
  long long sum = 1LL << (long long)k;
  if (s == x) sum -= 2;
  printf("%I64d\n", sum);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
long long s, x;
int main() {
  scanf("%I64d%I64d", &s, &x);
  long long cur = 0;
  for (long long i = 0; i < 63; ++i)
    if (x >> i & 1) cur |= 1LL << i;
  cur = s - cur;
  if (cur % 2 != 0 || ((cur / 2) & x) != 0) {
    puts("0");
    return 0;
  }
  cur /= 2;
  int k = 0;
  for (int i = 0; i < 64; ++i)
    if (x >> i & 1) ++k;
  long long sum = 1LL << (long long)k;
  if (s == x) sum -= 2;
  printf("%I64d\n", sum);
  return 0;
}

```