#include <bits/stdc++.h>
using namespace std;
long long int s, x, k;
long long int dp[43][2 + 3];
long long int cdp(int i, int b) {
  if (i == k) return ((b == 0) ? 1 : 0);
  if (dp[i][b] != -1) return dp[i][b];
  long long int ans = 0;
  for (int d1 = 0; d1 <= 1; d1++)
    for (int d2 = 0; d2 <= 1; d2++) {
      int si = 0, xi = 0;
      if (s & (1LL << i)) si = 1;
      if (x & (1LL << i)) xi = 1;
      if ((d1 + d2 + b) % 2 == si && (d1 ^ d2 == xi))
        ans += cdp(i + 1, (d1 + d2 + b) / 2);
    }
  return dp[i][b] = ans;
}
int main() {
  cin >> s >> x;
  if (x > s) {
    cout << 0 << endl;
    return 0;
  }
  for (long long int i = 0; i <= 40; i++)
    if (s & (1LL << i)) k = i;
  k++;
  for (long long int i = 0; i < k; i++)
    for (int j = 0; j < 2; j++) dp[i][j] = -1;
  long long int ans = cdp(0, 0);
  if (s == x) ans -= 2;
  cout << ans << 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;
long long int s, x, k;
long long int dp[43][2 + 3];
long long int cdp(int i, int b) {
  if (i == k) return ((b == 0) ? 1 : 0);
  if (dp[i][b] != -1) return dp[i][b];
  long long int ans = 0;
  for (int d1 = 0; d1 <= 1; d1++)
    for (int d2 = 0; d2 <= 1; d2++) {
      int si = 0, xi = 0;
      if (s & (1LL << i)) si = 1;
      if (x & (1LL << i)) xi = 1;
      if ((d1 + d2 + b) % 2 == si && (d1 ^ d2 == xi))
        ans += cdp(i + 1, (d1 + d2 + b) / 2);
    }
  return dp[i][b] = ans;
}
int main() {
  cin >> s >> x;
  if (x > s) {
    cout << 0 << endl;
    return 0;
  }
  for (long long int i = 0; i <= 40; i++)
    if (s & (1LL << i)) k = i;
  k++;
  for (long long int i = 0; i < k; i++)
    for (int j = 0; j < 2; j++) dp[i][j] = -1;
  long long int ans = cdp(0, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
}

```