#include <bits/stdc++.h>
using namespace std;
const int inf = (int)2 * 1e9;
const int MOD = (int)1e9 + 7;
const int p = 239;
const int MAX_N = 1000;
const long double pi = 3.1415926;
const long double eps = 1e-6;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  int a[41], b[41];
  for (int i = 0; i < 41; i++) a[i] = (s >> i) & 1;
  for (int i = 0; i < 41; i++) b[i] = (x >> i) & 1;
  long long dp[41][2];
  for (int i = 0; i < 41; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
  }
  if (a[0] == 1) {
    if (b[0] == 0) {
      dp[0][0] = 0;
      dp[0][1] = 0;
    } else {
      dp[0][0] = 2;
      dp[0][1] = 0;
    }
  } else {
    if (b[0] == 0) {
      dp[0][0] = 1;
      dp[0][1] = 1;
    } else {
      dp[0][0] = 0;
      dp[0][1] = 0;
    }
  }
  for (int i = 1; i < 41; i++)
    if (a[i] == 1) {
      if (b[i] == 0) {
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = dp[i - 1][1];
      } else {
        dp[i][0] = dp[i - 1][0] * 2;
        dp[i][1] = 0;
      }
    } else {
      if (b[i] == 0) {
        dp[i][0] = dp[i - 1][0];
        dp[i][1] = dp[i - 1][0];
      } else {
        dp[i][0] = 0;
        dp[i][1] = dp[i - 1][1] * 2;
      }
    }
  if (s != x)
    cout << dp[40][0];
  else
    cout << dp[40][0] - 2;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 inf = (int)2 * 1e9;
const int MOD = (int)1e9 + 7;
const int p = 239;
const int MAX_N = 1000;
const long double pi = 3.1415926;
const long double eps = 1e-6;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  int a[41], b[41];
  for (int i = 0; i < 41; i++) a[i] = (s >> i) & 1;
  for (int i = 0; i < 41; i++) b[i] = (x >> i) & 1;
  long long dp[41][2];
  for (int i = 0; i < 41; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
  }
  if (a[0] == 1) {
    if (b[0] == 0) {
      dp[0][0] = 0;
      dp[0][1] = 0;
    } else {
      dp[0][0] = 2;
      dp[0][1] = 0;
    }
  } else {
    if (b[0] == 0) {
      dp[0][0] = 1;
      dp[0][1] = 1;
    } else {
      dp[0][0] = 0;
      dp[0][1] = 0;
    }
  }
  for (int i = 1; i < 41; i++)
    if (a[i] == 1) {
      if (b[i] == 0) {
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = dp[i - 1][1];
      } else {
        dp[i][0] = dp[i - 1][0] * 2;
        dp[i][1] = 0;
      }
    } else {
      if (b[i] == 0) {
        dp[i][0] = dp[i - 1][0];
        dp[i][1] = dp[i - 1][0];
      } else {
        dp[i][0] = 0;
        dp[i][1] = dp[i - 1][1] * 2;
      }
    }
  if (s != x)
    cout << dp[40][0];
  else
    cout << dp[40][0] - 2;
  return 0;
}

```