#include <bits/stdc++.h>
using namespace std;
const int MAXN = 500005;
const int INF = 1000000000;
pair<int, int> deslocamento(int carry, int sum, int xorr) {
  if (xorr == 0) {
    if (sum == 0) {
      if (carry == 0) {
        return make_pair(1, 0);
      } else {
        return make_pair(1, 0);
      }
    } else {
      if (carry == 0) {
        return make_pair(1, 1);
      } else {
        return make_pair(1, 1);
      }
    }
  } else {
    if (sum == 0) {
      if (carry == 0) {
        return make_pair(0, 0);
      } else {
        return make_pair(2, 1);
      }
    } else {
      if (carry == 0) {
        return make_pair(2, 0);
      } else {
        return make_pair(0, 0);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  while (cin >> a >> b) {
    long long ans = 1;
    int carry = 0;
    for (int i = 58; i >= 0; i--) {
      pair<int, int> tmp = deslocamento(carry, (a >> i) & 1, (b >> i) & 1);
      ans *= tmp.first;
      carry = tmp.second;
    }
    if (carry == 1) ans = 0;
    if (ans > 1 && a == b) ans -= 2;
    cout << ans << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAXN = 500005;
const int INF = 1000000000;
pair<int, int> deslocamento(int carry, int sum, int xorr) {
  if (xorr == 0) {
    if (sum == 0) {
      if (carry == 0) {
        return make_pair(1, 0);
      } else {
        return make_pair(1, 0);
      }
    } else {
      if (carry == 0) {
        return make_pair(1, 1);
      } else {
        return make_pair(1, 1);
      }
    }
  } else {
    if (sum == 0) {
      if (carry == 0) {
        return make_pair(0, 0);
      } else {
        return make_pair(2, 1);
      }
    } else {
      if (carry == 0) {
        return make_pair(2, 0);
      } else {
        return make_pair(0, 0);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  while (cin >> a >> b) {
    long long ans = 1;
    int carry = 0;
    for (int i = 58; i >= 0; i--) {
      pair<int, int> tmp = deslocamento(carry, (a >> i) & 1, (b >> i) & 1);
      ans *= tmp.first;
      carry = tmp.second;
    }
    if (carry == 1) ans = 0;
    if (ans > 1 && a == b) ans -= 2;
    cout << ans << endl;
  }
}

```