#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
map<pair<long long, long long>, long long> dp;
long long f(long long s, long long x) {
  if (dp.count(make_pair(s, x))) return dp[make_pair(s, x)];
  if (s <= 8) {
    int an = 0;
    for (int i = 0; i < (s + 1); ++i) {
      if ((i ^ (s - i)) == x) an++;
    }
    return dp[make_pair(s, x)] = an;
  }
  if (x & 1) {
    if (s & 1) return dp[make_pair(s, x)] = 2 * f(s >> 1, x >> 1);
    return dp[make_pair(s, x)] = 0;
  } else {
    if (s & 1) return dp[make_pair(s, x)] = 0;
    return dp[make_pair(s, x)] = f(s >> 1, x >> 1) + f((s >> 1) - 1, x >> 1);
  }
}
int main() {
  long long s, x;
  cin >> s >> x;
  printf("%I64d\n", f(s, x) - (s == x) * 2);
  return 0;
}

### 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;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
map<pair<long long, long long>, long long> dp;
long long f(long long s, long long x) {
  if (dp.count(make_pair(s, x))) return dp[make_pair(s, x)];
  if (s <= 8) {
    int an = 0;
    for (int i = 0; i < (s + 1); ++i) {
      if ((i ^ (s - i)) == x) an++;
    }
    return dp[make_pair(s, x)] = an;
  }
  if (x & 1) {
    if (s & 1) return dp[make_pair(s, x)] = 2 * f(s >> 1, x >> 1);
    return dp[make_pair(s, x)] = 0;
  } else {
    if (s & 1) return dp[make_pair(s, x)] = 0;
    return dp[make_pair(s, x)] = f(s >> 1, x >> 1) + f((s >> 1) - 1, x >> 1);
  }
}
int main() {
  long long s, x;
  cin >> s >> x;
  printf("%I64d\n", f(s, x) - (s == x) * 2);
  return 0;
}

```