#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int inf = 1e9;
const int mod = 1e9 + 7;
int n;
long long dp[maxn][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long s, x;
  cin >> s >> x;
  dp[50][0] = 1;
  for (long long i = 49; i >= 0; i--) {
    if (x & (1ll << i)) {
      if (s & (1ll << i)) {
        dp[i][0] = 2 * dp[i + 1][0];
      } else
        dp[i][1] = 2 * dp[i + 1][1];
    } else {
      if (s & (1ll << i)) {
        dp[i][1] = dp[i + 1][1] + dp[i + 1][0];
      } else {
        dp[i][0] = dp[i + 1][1] + dp[i + 1][0];
      }
    }
  }
  if (x == s) dp[0][0] -= 2;
  cout << dp[0][0] << endl;
}

### Prompt

In Cpp, your task is to solve 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;
const int maxn = 1e5 + 5;
const int inf = 1e9;
const int mod = 1e9 + 7;
int n;
long long dp[maxn][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long s, x;
  cin >> s >> x;
  dp[50][0] = 1;
  for (long long i = 49; i >= 0; i--) {
    if (x & (1ll << i)) {
      if (s & (1ll << i)) {
        dp[i][0] = 2 * dp[i + 1][0];
      } else
        dp[i][1] = 2 * dp[i + 1][1];
    } else {
      if (s & (1ll << i)) {
        dp[i][1] = dp[i + 1][1] + dp[i + 1][0];
      } else {
        dp[i][0] = dp[i + 1][1] + dp[i + 1][0];
      }
    }
  }
  if (x == s) dp[0][0] -= 2;
  cout << dp[0][0] << endl;
}

```