#include <bits/stdc++.h>
using namespace std;
void read_file(bool outToFile = true) {}
long long s, x;
long long memo[64][100];
long long f(int i = 59, long long a = 0, long long b = 0, long long sum = 0) {
  if (i == -1) return sum == s;
  if (x & (1LL << i)) {
    return f(i - 1, a, b | (1LL << i), sum + (1LL << i));
  } else {
    return f(i - 1, a, b) + f(i + 1, a | (1LL << i), b | (1LL << i));
    return f(i + 1);
  }
}
long long rec(int i = 0, int cary = 0) {
  if (i == 62) return cary == 0;
  long long &ret = memo[i][cary];
  if (ret != -1) return ret;
  ret = 0;
  if ((x & (1LL << i)) != 0) {
    if ((cary + 1) % 2 == ((s & (1LL << i)) != 0))
      ret += 2 * rec(i + 1, (cary + 1) / 2);
  } else {
    if (cary % 2 == ((s & (1LL << i)) != 0)) {
      ret += rec(i + 1, cary / 2);
      ret += rec(i + 1, cary / 2 + 1);
    }
  }
  return ret;
}
int main() {
  read_file();
  while (scanf("%lld%lld\n", &s, &x) != EOF) {
    memset(memo, -1, sizeof memo);
    long long ans = rec();
    if (s == x) ans -= 2;
    printf("%lld\n", ans);
  }
}

### 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;
void read_file(bool outToFile = true) {}
long long s, x;
long long memo[64][100];
long long f(int i = 59, long long a = 0, long long b = 0, long long sum = 0) {
  if (i == -1) return sum == s;
  if (x & (1LL << i)) {
    return f(i - 1, a, b | (1LL << i), sum + (1LL << i));
  } else {
    return f(i - 1, a, b) + f(i + 1, a | (1LL << i), b | (1LL << i));
    return f(i + 1);
  }
}
long long rec(int i = 0, int cary = 0) {
  if (i == 62) return cary == 0;
  long long &ret = memo[i][cary];
  if (ret != -1) return ret;
  ret = 0;
  if ((x & (1LL << i)) != 0) {
    if ((cary + 1) % 2 == ((s & (1LL << i)) != 0))
      ret += 2 * rec(i + 1, (cary + 1) / 2);
  } else {
    if (cary % 2 == ((s & (1LL << i)) != 0)) {
      ret += rec(i + 1, cary / 2);
      ret += rec(i + 1, cary / 2 + 1);
    }
  }
  return ret;
}
int main() {
  read_file();
  while (scanf("%lld%lld\n", &s, &x) != EOF) {
    memset(memo, -1, sizeof memo);
    long long ans = rec();
    if (s == x) ans -= 2;
    printf("%lld\n", ans);
  }
}

```