#include <bits/stdc++.h>
int main() {
  long long int a, b, sum, i, xo, bit[75] = {0};
  scanf("%lld%lld", &sum, &xo);
  int car = 0, f = 0;
  for (i = 60; i >= 0; i--) {
    int sbit = 0, xbit = 0;
    if (sum & ((long long int)1 << i)) sbit = 1;
    if (xo & ((long long int)1 << i)) xbit = 1;
    if (sbit == 1 && xbit == 0) {
      if (car == 1) {
        bit[i] = 2;
        car = 1;
      } else {
        car = 1;
        bit[i] = 0;
      }
    } else if (sbit == 1 && xbit == 1) {
      if (car == 1) f = 1;
      car = 0;
      bit[i] = 1;
    } else if (sbit == 0 && xbit == 0) {
      if (car == 0) {
        bit[i] = 0;
        car = 0;
      } else {
        car = 0;
        bit[i] = 2;
      }
    } else {
      if (car == 0) f = 1;
      car = 1;
      bit[i] = 1;
    }
  }
  long long int ans = 1;
  if (car == 1 || f == 1) {
    printf("0");
    return 0;
  }
  int two = 0, one = 0;
  for (i = 0; i < 60; i++) {
    if (bit[i] == 2) two++;
    if (bit[i] == 1) {
      one++;
      ans = ans * (long long int)2;
    }
  }
  if (two == 0) ans = ans - 2;
  printf("%lld", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
int main() {
  long long int a, b, sum, i, xo, bit[75] = {0};
  scanf("%lld%lld", &sum, &xo);
  int car = 0, f = 0;
  for (i = 60; i >= 0; i--) {
    int sbit = 0, xbit = 0;
    if (sum & ((long long int)1 << i)) sbit = 1;
    if (xo & ((long long int)1 << i)) xbit = 1;
    if (sbit == 1 && xbit == 0) {
      if (car == 1) {
        bit[i] = 2;
        car = 1;
      } else {
        car = 1;
        bit[i] = 0;
      }
    } else if (sbit == 1 && xbit == 1) {
      if (car == 1) f = 1;
      car = 0;
      bit[i] = 1;
    } else if (sbit == 0 && xbit == 0) {
      if (car == 0) {
        bit[i] = 0;
        car = 0;
      } else {
        car = 0;
        bit[i] = 2;
      }
    } else {
      if (car == 0) f = 1;
      car = 1;
      bit[i] = 1;
    }
  }
  long long int ans = 1;
  if (car == 1 || f == 1) {
    printf("0");
    return 0;
  }
  int two = 0, one = 0;
  for (i = 0; i < 60; i++) {
    if (bit[i] == 2) two++;
    if (bit[i] == 1) {
      one++;
      ans = ans * (long long int)2;
    }
  }
  if (two == 0) ans = ans - 2;
  printf("%lld", ans);
  return 0;
}

```