#include <bits/stdc++.h>
using namespace std;
vector<long long> to_d(long long x) {
  long long y = x;
  vector<long long> a(50, 0);
  long long i = 1;
  while (y != 0) {
    a[i] = y % 2;
    i++;
    y /= 2;
  }
  return a;
}
vector<long long> a, b;
long long ans = 0;
int main() {
  long long s, x;
  cin >> s >> x;
  if (x == 0) {
    cout << 1 - (s % 2) << endl;
    return 0;
  }
  a = to_d(s);
  b = to_d(x);
  long long dp[50][2];
  for (long long i = 0; i < 50; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
  }
  dp[0][0] = 1;
  for (long long i = 1; i < 50; i++) {
    if ((a[i] == 0) && (b[i] == 0)) {
      dp[i][0] += dp[i - 1][0];
      dp[i][1] += dp[i - 1][0];
    }
    if ((a[i] == 1) && (b[i] == 1)) {
      dp[i][0] += 2 * dp[i - 1][0];
    }
    if ((a[i] == 1) && (b[i] == 0)) {
      dp[i][1] += dp[i - 1][1];
      dp[i][0] += dp[i - 1][1];
    }
    if ((a[i] == 0) && (b[i] == 1)) {
      dp[i][1] += 2 * dp[i - 1][1];
    }
  }
  long long ans = dp[47][0];
  if (x == s) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
vector<long long> to_d(long long x) {
  long long y = x;
  vector<long long> a(50, 0);
  long long i = 1;
  while (y != 0) {
    a[i] = y % 2;
    i++;
    y /= 2;
  }
  return a;
}
vector<long long> a, b;
long long ans = 0;
int main() {
  long long s, x;
  cin >> s >> x;
  if (x == 0) {
    cout << 1 - (s % 2) << endl;
    return 0;
  }
  a = to_d(s);
  b = to_d(x);
  long long dp[50][2];
  for (long long i = 0; i < 50; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
  }
  dp[0][0] = 1;
  for (long long i = 1; i < 50; i++) {
    if ((a[i] == 0) && (b[i] == 0)) {
      dp[i][0] += dp[i - 1][0];
      dp[i][1] += dp[i - 1][0];
    }
    if ((a[i] == 1) && (b[i] == 1)) {
      dp[i][0] += 2 * dp[i - 1][0];
    }
    if ((a[i] == 1) && (b[i] == 0)) {
      dp[i][1] += dp[i - 1][1];
      dp[i][0] += dp[i - 1][1];
    }
    if ((a[i] == 0) && (b[i] == 1)) {
      dp[i][1] += 2 * dp[i - 1][1];
    }
  }
  long long ans = dp[47][0];
  if (x == s) ans -= 2;
  cout << ans;
  return 0;
}

```