#include <bits/stdc++.h>
using namespace std;
vector<int> G[2 * 100000 + 5];
signed long long int S, X;
int fdp[60][3];
signed long long int dp[60][3];
bool checkBit(signed long long int x, int b) { return (x & (1LL << b)); }
signed long long int rec(int b, int c) {
  if (b == 60) {
    if (c == 0) return 1;
    return 0;
  }
  if (fdp[b][c]) return dp[b][c];
  fdp[b][c] = 1;
  dp[b][c] = 0;
  for (auto m : {0, 1}) {
    for (auto n : {0, 1}) {
      bool s = checkBit(m + n + c, 0);
      bool new_c = checkBit(m + n + c, 1);
      bool x = m ^ n;
      if (s == checkBit(S, b) and x == checkBit(X, b)) {
        dp[b][c] += rec(b + 1, new_c);
      }
    }
  }
  return dp[b][c];
}
int main() {
  cin >> S >> X;
  cout << rec(0, 0) - (S == X ? 2 : 0);
}

### 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;
vector<int> G[2 * 100000 + 5];
signed long long int S, X;
int fdp[60][3];
signed long long int dp[60][3];
bool checkBit(signed long long int x, int b) { return (x & (1LL << b)); }
signed long long int rec(int b, int c) {
  if (b == 60) {
    if (c == 0) return 1;
    return 0;
  }
  if (fdp[b][c]) return dp[b][c];
  fdp[b][c] = 1;
  dp[b][c] = 0;
  for (auto m : {0, 1}) {
    for (auto n : {0, 1}) {
      bool s = checkBit(m + n + c, 0);
      bool new_c = checkBit(m + n + c, 1);
      bool x = m ^ n;
      if (s == checkBit(S, b) and x == checkBit(X, b)) {
        dp[b][c] += rec(b + 1, new_c);
      }
    }
  }
  return dp[b][c];
}
int main() {
  cin >> S >> X;
  cout << rec(0, 0) - (S == X ? 2 : 0);
}

```