#include <bits/stdc++.h>
using namespace std;
string scan();
const int mod = 1000000007;
const int MAXN = 10010;
int sta[MAXN], stb[MAXN];
long long s, x;
long long memo[MAXN][2];
long long bt(int i, int rem) {
  if (i > 52 && rem == 0) return 1;
  if (i == 60) return 0;
  long long &ret = memo[i][rem];
  if (ret != -1) return ret;
  ret = 0;
  if (sta[i] == 1 && ((rem + 1) % 2) == stb[i])
    return ret = 2 * bt(i + 1, (rem + 1) / 2);
  if (sta[i] == 1) return rem = 0;
  if ((rem + 2) % 2 == stb[i]) ret += bt(i + 1, (rem + 2) / 2);
  if (rem % 2 == stb[i]) ret += bt(i + 1, 0);
  return ret;
}
int main() {
  memset(memo, -1, sizeof memo);
  long long a, b;
  cin >> a >> b;
  s = a, x = b;
  for (int i = 0; a; a /= 2, i++) stb[i] = a % 2;
  for (int i = 0; b; b /= 2, i++) sta[i] = b % 2;
  cout << bt(0, 0) - (s == x ? 2 : 0) << endl;
}

### Prompt

Your challenge is to write 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;
string scan();
const int mod = 1000000007;
const int MAXN = 10010;
int sta[MAXN], stb[MAXN];
long long s, x;
long long memo[MAXN][2];
long long bt(int i, int rem) {
  if (i > 52 && rem == 0) return 1;
  if (i == 60) return 0;
  long long &ret = memo[i][rem];
  if (ret != -1) return ret;
  ret = 0;
  if (sta[i] == 1 && ((rem + 1) % 2) == stb[i])
    return ret = 2 * bt(i + 1, (rem + 1) / 2);
  if (sta[i] == 1) return rem = 0;
  if ((rem + 2) % 2 == stb[i]) ret += bt(i + 1, (rem + 2) / 2);
  if (rem % 2 == stb[i]) ret += bt(i + 1, 0);
  return ret;
}
int main() {
  memset(memo, -1, sizeof memo);
  long long a, b;
  cin >> a >> b;
  s = a, x = b;
  for (int i = 0; a; a /= 2, i++) stb[i] = a % 2;
  for (int i = 0; b; b /= 2, i++) sta[i] = b % 2;
  cout << bt(0, 0) - (s == x ? 2 : 0) << endl;
}

```