#include <bits/stdc++.h>
using namespace std;
void smain();
int main() {
  smain();
  return 0;
}
long long s, x;
long long naive() {
  long long res = 0;
  for (long long i = 0; i < s + 1; i++) {
    long long j = s - i;
    if (i == 0 || j == 0) continue;
    if ((i ^ j) == x) res += 1;
  }
  return res;
}
long long cnt[1 << 21];
long long solve() {
  long long mm = 1 << 20;
  memset(cnt, 0, sizeof(cnt));
  for (long long mask = 0; mask < mm; mask++) {
    long long comp = (x & (mm - 1)) ^ mask;
    cnt[mask + comp] += 1;
  }
  long long res = 0;
  for (long long mask = 0; mask < mm; mask++) {
    long long comp = ((x >> 20) & (mm - 1)) ^ mask;
    long long val = (comp << 20) + (mask << 20);
    long long rem = s - val;
    if (rem >= 0 && rem < (1 << 21)) {
      res += cnt[rem];
    }
  }
  if (s == x) res -= 2;
  return res;
}
void smain() {
  for (; cin >> s >> x;) {
    cout << solve() << endl;
  }
}

### Prompt

Your task is to create 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;
void smain();
int main() {
  smain();
  return 0;
}
long long s, x;
long long naive() {
  long long res = 0;
  for (long long i = 0; i < s + 1; i++) {
    long long j = s - i;
    if (i == 0 || j == 0) continue;
    if ((i ^ j) == x) res += 1;
  }
  return res;
}
long long cnt[1 << 21];
long long solve() {
  long long mm = 1 << 20;
  memset(cnt, 0, sizeof(cnt));
  for (long long mask = 0; mask < mm; mask++) {
    long long comp = (x & (mm - 1)) ^ mask;
    cnt[mask + comp] += 1;
  }
  long long res = 0;
  for (long long mask = 0; mask < mm; mask++) {
    long long comp = ((x >> 20) & (mm - 1)) ^ mask;
    long long val = (comp << 20) + (mask << 20);
    long long rem = s - val;
    if (rem >= 0 && rem < (1 << 21)) {
      res += cnt[rem];
    }
  }
  if (s == x) res -= 2;
  return res;
}
void smain() {
  for (; cin >> s >> x;) {
    cout << solve() << endl;
  }
}

```