#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000009ll;
const long long INF = 9223372036854775807ll;
long long power(long long x, long long y, long long z) {
  long long t = 1;
  while (y > 0) {
    if (y % 2) {
      t = (t * x) % z;
    }
    x = (x * x) % z;
  }
  return t;
}
long long p, s, x;
int A[100], ans = 0;
long long B[100], k, p1;
int main() {
  int i, sz = 0;
  scanf("%lld%lld", &s, &x);
  k = s - x;
  if (k & 1) {
    printf("0\n");
    return 0;
  }
  k /= 2;
  B[0] = 1;
  for (i = 1; i <= 62; i++) B[i] = B[i - 1] * 2ll;
  for (i = 61; i >= 0; i--) {
    p = x & B[i];
    p1 = k & B[i];
    if (p > 0 && p == p1) {
      printf("0\n");
      return 0;
    }
    if (p > 0) ans++;
  }
  if (ans == 0) {
    if (s % 2 != 0)
      printf("0\n");
    else
      printf("1\n");
  } else {
    if (k == 0)
      printf("%lld\n", B[ans] - 2);
    else
      printf("%lld\n", B[ans]);
  }
  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;
const long long MOD = 1000000009ll;
const long long INF = 9223372036854775807ll;
long long power(long long x, long long y, long long z) {
  long long t = 1;
  while (y > 0) {
    if (y % 2) {
      t = (t * x) % z;
    }
    x = (x * x) % z;
  }
  return t;
}
long long p, s, x;
int A[100], ans = 0;
long long B[100], k, p1;
int main() {
  int i, sz = 0;
  scanf("%lld%lld", &s, &x);
  k = s - x;
  if (k & 1) {
    printf("0\n");
    return 0;
  }
  k /= 2;
  B[0] = 1;
  for (i = 1; i <= 62; i++) B[i] = B[i - 1] * 2ll;
  for (i = 61; i >= 0; i--) {
    p = x & B[i];
    p1 = k & B[i];
    if (p > 0 && p == p1) {
      printf("0\n");
      return 0;
    }
    if (p > 0) ans++;
  }
  if (ans == 0) {
    if (s % 2 != 0)
      printf("0\n");
    else
      printf("1\n");
  } else {
    if (k == 0)
      printf("%lld\n", B[ans] - 2);
    else
      printf("%lld\n", B[ans]);
  }
  return 0;
}

```