#include <bits/stdc++.h>
using namespace std;
bool res1[100], res2[100];
int main() {
  long long s = 0, x = 0;
  scanf("%lld%lld", &s, &x);
  long long i = 1;
  long long tmp = x;
  long long cnt = 1;
  long long And = (s - x) / 2;
  long long tmp2 = And;
  bool flag = 0;
  bool flag2 = 0;
  while (x || And) {
    res1[i++] = x % 2;
    if (x % 2 == 1) {
      cnt = cnt * 2;
    }
    if ((x % 2 == 0) && (And % 2 != 0)) {
      flag = 1;
    }
    if ((x % 2 == 1) && (And % 2 != 0)) {
      flag2 = 1;
    }
    x = x / 2;
    And = And / 2;
  }
  if (flag == 0) {
    cnt = cnt - 2;
  }
  if (((s - tmp) % 2 == 1) || (s - tmp < 0) || flag2) {
    puts("0");
  } else {
    printf("%lld\n", cnt);
  }
}

### Prompt

Generate 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>
using namespace std;
bool res1[100], res2[100];
int main() {
  long long s = 0, x = 0;
  scanf("%lld%lld", &s, &x);
  long long i = 1;
  long long tmp = x;
  long long cnt = 1;
  long long And = (s - x) / 2;
  long long tmp2 = And;
  bool flag = 0;
  bool flag2 = 0;
  while (x || And) {
    res1[i++] = x % 2;
    if (x % 2 == 1) {
      cnt = cnt * 2;
    }
    if ((x % 2 == 0) && (And % 2 != 0)) {
      flag = 1;
    }
    if ((x % 2 == 1) && (And % 2 != 0)) {
      flag2 = 1;
    }
    x = x / 2;
    And = And / 2;
  }
  if (flag == 0) {
    cnt = cnt - 2;
  }
  if (((s - tmp) % 2 == 1) || (s - tmp < 0) || flag2) {
    puts("0");
  } else {
    printf("%lld\n", cnt);
  }
}

```