#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long dp[55][2];
long long f(long long idx, long long carry) {
  if ((1LL << idx) > s) return carry ^ 1LL;
  if (~dp[idx][carry]) return dp[idx][carry];
  long long res = 0;
  if ((x >> idx) & 1LL) {
    if ((s >> idx) & 1LL) {
      if (!carry) res = 2LL * f(idx + 1, 0);
    } else {
      if (carry) res = 2LL * f(idx + 1, 1);
    }
  } else {
    if ((s >> idx) & 1LL) {
      if (carry) res = f(idx + 1, 0) + f(idx + 1, 1);
    } else {
      if (!carry) res = f(idx + 1, 0) + f(idx + 1, 1);
    }
  }
  return dp[idx][carry] = res;
}
int32_t main() {
  cin >> s >> x;
  if (s < x) {
    cout << 0 << endl;
    return 0;
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0) - 2LL * (s == x) << 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 s, x;
long long dp[55][2];
long long f(long long idx, long long carry) {
  if ((1LL << idx) > s) return carry ^ 1LL;
  if (~dp[idx][carry]) return dp[idx][carry];
  long long res = 0;
  if ((x >> idx) & 1LL) {
    if ((s >> idx) & 1LL) {
      if (!carry) res = 2LL * f(idx + 1, 0);
    } else {
      if (carry) res = 2LL * f(idx + 1, 1);
    }
  } else {
    if ((s >> idx) & 1LL) {
      if (carry) res = f(idx + 1, 0) + f(idx + 1, 1);
    } else {
      if (!carry) res = f(idx + 1, 0) + f(idx + 1, 1);
    }
  }
  return dp[idx][carry] = res;
}
int32_t main() {
  cin >> s >> x;
  if (s < x) {
    cout << 0 << endl;
    return 0;
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0) - 2LL * (s == x) << endl;
}

```