#include <bits/stdc++.h>
using namespace std;
long long int em(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return em((x % 1000000007 * x % 1000000007) % 1000000007, n / 2);
  else
    return (x % 1000000007 *
            em((x % 1000000007 * x % 1000000007) % 1000000007, (n - 1) / 2)) %
           1000000007;
}
long long int su, x, li;
long long int dp[45][3];
long long int fun(long long int n, long long int c) {
  if (n == -1 && c == 0) return 1;
  if (n == -1 && c == 1) return 0;
  if (dp[n][c] != -1) return dp[n][c];
  long long int ans = 0;
  long long int c1, c2;
  c1 = (su & (1ll << n));
  c2 = (x & (1ll << n));
  if (c == 0) {
    if (c1 == 0 && c2 == 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 == 0)
      ans += fun(n - 1, 1);
    else if (c1 == 0 && c2 != 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 != 0)
      ans += fun(n - 1, 0);
  } else {
    if (c1 == 0 && c2 == 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 == 0)
      ans += fun(n - 1, 1);
    else if (c1 == 0 && c2 != 0)
      ans += fun(n - 1, 1);
    else if (c1 != 0 && c2 != 0)
      ans += 0;
  }
  return dp[n][c] = ans;
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k;
  cin >> su >> x;
  dp[0][0] = 1;
  for (i = 0; i <= 43; i++) {
    long long int c1 = (su & (1ll << i));
    long long int c2 = (x & (1ll << i));
    if (c1 == 0 && c2 == 0) {
      dp[i + 1][1] += dp[i][0];
      dp[i + 1][0] += dp[i][0];
    } else if (c1 == 0 && c2 != 0) {
      dp[i + 1][1] += 2 * dp[i][1];
    } else if (c1 != 0 && c2 == 0) {
      dp[i + 1][0] += dp[i][1];
      dp[i + 1][1] += dp[i][1];
    } else if (c1 != 0 && c2 != 0) {
      dp[i + 1][0] += 2 * dp[i][0];
    }
  }
  if (su == x)
    cout << dp[43][0] - 2;
  else
    cout << dp[43][0];
}

### Prompt

Create a solution in Cpp for 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 em(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return em((x % 1000000007 * x % 1000000007) % 1000000007, n / 2);
  else
    return (x % 1000000007 *
            em((x % 1000000007 * x % 1000000007) % 1000000007, (n - 1) / 2)) %
           1000000007;
}
long long int su, x, li;
long long int dp[45][3];
long long int fun(long long int n, long long int c) {
  if (n == -1 && c == 0) return 1;
  if (n == -1 && c == 1) return 0;
  if (dp[n][c] != -1) return dp[n][c];
  long long int ans = 0;
  long long int c1, c2;
  c1 = (su & (1ll << n));
  c2 = (x & (1ll << n));
  if (c == 0) {
    if (c1 == 0 && c2 == 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 == 0)
      ans += fun(n - 1, 1);
    else if (c1 == 0 && c2 != 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 != 0)
      ans += fun(n - 1, 0);
  } else {
    if (c1 == 0 && c2 == 0)
      ans += fun(n - 1, 0);
    else if (c1 != 0 && c2 == 0)
      ans += fun(n - 1, 1);
    else if (c1 == 0 && c2 != 0)
      ans += fun(n - 1, 1);
    else if (c1 != 0 && c2 != 0)
      ans += 0;
  }
  return dp[n][c] = ans;
}
signed main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k;
  cin >> su >> x;
  dp[0][0] = 1;
  for (i = 0; i <= 43; i++) {
    long long int c1 = (su & (1ll << i));
    long long int c2 = (x & (1ll << i));
    if (c1 == 0 && c2 == 0) {
      dp[i + 1][1] += dp[i][0];
      dp[i + 1][0] += dp[i][0];
    } else if (c1 == 0 && c2 != 0) {
      dp[i + 1][1] += 2 * dp[i][1];
    } else if (c1 != 0 && c2 == 0) {
      dp[i + 1][0] += dp[i][1];
      dp[i + 1][1] += dp[i][1];
    } else if (c1 != 0 && c2 != 0) {
      dp[i + 1][0] += 2 * dp[i][0];
    }
  }
  if (su == x)
    cout << dp[43][0] - 2;
  else
    cout << dp[43][0];
}

```