#include <bits/stdc++.h>
using namespace std;
long long dp[41][2][2][2];
long long doit(int ix, int cr, int ap, int bp, long long a, long long b) {
  if (ix == 0) {
    if (a & 1) {
      if (cr) return 0;
      int res = 0;
      if (b & 1) {
        if (ap) res++;
        if (bp) res++;
      }
      return res;
    } else {
      if (cr && !(b & 1)) return 1;
      if (cr == 0 && !(b & 1) && ap && bp) return 1;
      return 0;
    }
  }
  long long &ret = dp[ix][cr][ap][bp];
  if (ret != -1) return ret;
  ret = 0;
  long long bit = (1LL << ix);
  if (a & bit) {
    if (cr) {
      if (!(b & bit)) {
        ret += doit(ix - 1, 1, 1, 1, a, b);
      }
    } else {
      if (b & bit) {
        ret += doit(ix - 1, 0, 1, bp, a, b);
        ret += doit(ix - 1, 0, ap, 1, a, b);
      }
      if (!(b & bit)) {
        ret += doit(ix - 1, 1, ap, bp, a, b);
      }
    }
  } else {
    if (cr) {
      if (b & bit) {
        ret += doit(ix - 1, 1, 1, bp, a, b);
        ret += doit(ix - 1, 1, ap, 1, a, b);
      }
      if (!(b & bit)) {
        ret += doit(ix - 1, 0, 1, 1, a, b);
      }
    } else {
      if (!(b & bit)) {
        ret += doit(ix - 1, 0, ap, bp, a, b);
      }
    }
  }
  return ret;
}
int main() {
  long long a, b;
  cin >> a >> b;
  memset(dp, -1, sizeof(dp));
  cout << doit(40, 0, 0, 0, a, b) << endl;
  return 0;
}

### Prompt

Please formulate 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 dp[41][2][2][2];
long long doit(int ix, int cr, int ap, int bp, long long a, long long b) {
  if (ix == 0) {
    if (a & 1) {
      if (cr) return 0;
      int res = 0;
      if (b & 1) {
        if (ap) res++;
        if (bp) res++;
      }
      return res;
    } else {
      if (cr && !(b & 1)) return 1;
      if (cr == 0 && !(b & 1) && ap && bp) return 1;
      return 0;
    }
  }
  long long &ret = dp[ix][cr][ap][bp];
  if (ret != -1) return ret;
  ret = 0;
  long long bit = (1LL << ix);
  if (a & bit) {
    if (cr) {
      if (!(b & bit)) {
        ret += doit(ix - 1, 1, 1, 1, a, b);
      }
    } else {
      if (b & bit) {
        ret += doit(ix - 1, 0, 1, bp, a, b);
        ret += doit(ix - 1, 0, ap, 1, a, b);
      }
      if (!(b & bit)) {
        ret += doit(ix - 1, 1, ap, bp, a, b);
      }
    }
  } else {
    if (cr) {
      if (b & bit) {
        ret += doit(ix - 1, 1, 1, bp, a, b);
        ret += doit(ix - 1, 1, ap, 1, a, b);
      }
      if (!(b & bit)) {
        ret += doit(ix - 1, 0, 1, 1, a, b);
      }
    } else {
      if (!(b & bit)) {
        ret += doit(ix - 1, 0, ap, bp, a, b);
      }
    }
  }
  return ret;
}
int main() {
  long long a, b;
  cin >> a >> b;
  memset(dp, -1, sizeof(dp));
  cout << doit(40, 0, 0, 0, a, b) << endl;
  return 0;
}

```