#include <bits/stdc++.h>
using namespace std;
int vis[60];
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  long long ans = 1;
  for (int i = 0; i < 60; i++) {
    if (x & (1ll << i)) {
      vis[i] = 1;
      s -= (1ll << i);
      ans = ans * 2;
    }
  }
  if (s % 2 != 0 || s < 0) {
    printf("0\n");
    return 0;
  }
  s /= 2;
  int flag = 1;
  for (int i = 0; i < 60; i++) {
    if (s & (1ll << i)) {
      if (vis[i]) flag = 0;
    }
  }
  if (!flag) {
    printf("0\n");
    return 0;
  }
  if (s == 0) {
    printf("%I64d\n", ans - 2);
    return 0;
  }
  printf("%I64d\n", ans);
}

### 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>
using namespace std;
int vis[60];
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  long long ans = 1;
  for (int i = 0; i < 60; i++) {
    if (x & (1ll << i)) {
      vis[i] = 1;
      s -= (1ll << i);
      ans = ans * 2;
    }
  }
  if (s % 2 != 0 || s < 0) {
    printf("0\n");
    return 0;
  }
  s /= 2;
  int flag = 1;
  for (int i = 0; i < 60; i++) {
    if (s & (1ll << i)) {
      if (vis[i]) flag = 0;
    }
  }
  if (!flag) {
    printf("0\n");
    return 0;
  }
  if (s == 0) {
    printf("%I64d\n", ans - 2);
    return 0;
  }
  printf("%I64d\n", ans);
}

```