#include <bits/stdc++.h>
using namespace std;
long long s1[110], x1[110];
int main() {
  memset(s1, 0, sizeof(s1));
  memset(x1, 0, sizeof(x1));
  long long s, x, i, j;
  scanf("%I64d %I64d", &s, &x);
  if (!x) {
    if (s % 2 == 0)
      printf("1");
    else
      printf("0");
    return 0;
  }
  long long t1 = s, t2 = x;
  long long yy = (t1 - t2);
  if (yy % 2 || t1 < t2) {
    printf("0");
    return 0;
  }
  yy /= 2;
  long long temp = 1;
  for (i = 0; i < 60; i++) {
    int x1 = (yy >> i) & 1;
    int x2 = (x >> i) & 1;
    if (x1 == 1 && x2 == 1) {
      printf("0");
      return 0;
    }
    if (x2 == 1) temp *= 2;
  }
  if (t1 == t2) {
    printf("%I64d", temp - 2);
  } else {
    printf("%I64d", temp);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long s1[110], x1[110];
int main() {
  memset(s1, 0, sizeof(s1));
  memset(x1, 0, sizeof(x1));
  long long s, x, i, j;
  scanf("%I64d %I64d", &s, &x);
  if (!x) {
    if (s % 2 == 0)
      printf("1");
    else
      printf("0");
    return 0;
  }
  long long t1 = s, t2 = x;
  long long yy = (t1 - t2);
  if (yy % 2 || t1 < t2) {
    printf("0");
    return 0;
  }
  yy /= 2;
  long long temp = 1;
  for (i = 0; i < 60; i++) {
    int x1 = (yy >> i) & 1;
    int x2 = (x >> i) & 1;
    if (x1 == 1 && x2 == 1) {
      printf("0");
      return 0;
    }
    if (x2 == 1) temp *= 2;
  }
  if (t1 == t2) {
    printf("%I64d", temp - 2);
  } else {
    printf("%I64d", temp);
  }
  return 0;
}

```