#include <bits/stdc++.h>
using namespace std;
long long s, x;
int n;
long long dp[64][2][2];
long long getBit(long long mask, long long curr) {
  return (mask >> curr) & 1LL;
}
long long solve(int curr, int c, int f) {
  if (curr == n + 1) return f && !c;
  long long& ret = dp[curr][c][f];
  if (ret != -1) return ret;
  ret = 0;
  if ((c + 1 + (getBit(x, curr) ^ 1)) % 2 == getBit(s, curr)) {
    ret += solve(curr + 1, (c + 1 + (getBit(x, curr) ^ 1)) >= 2, 1);
  }
  if ((c + 0 + (getBit(x, curr) ^ 0)) % 2 == getBit(s, curr)) {
    ret += solve(curr + 1, (c + 0 + (getBit(x, curr) ^ 0)) >= 2, f);
  }
  return ret;
}
int main(void) {
  cin >> s >> x;
  for (int i = 59; i >= 0; --i) {
    if (getBit(x, i) && !getBit(s, i)) {
      cout << 0 << "\n";
      return 0;
    } else if (getBit(s, i)) {
      n = i;
      break;
    }
  }
  memset(dp, -1, sizeof dp);
  long long ret = solve(0, 0, 0);
  if (x == s)
    cout << ret - 1 << "\n";
  else
    cout << ret << "\n";
  return 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;
long long s, x;
int n;
long long dp[64][2][2];
long long getBit(long long mask, long long curr) {
  return (mask >> curr) & 1LL;
}
long long solve(int curr, int c, int f) {
  if (curr == n + 1) return f && !c;
  long long& ret = dp[curr][c][f];
  if (ret != -1) return ret;
  ret = 0;
  if ((c + 1 + (getBit(x, curr) ^ 1)) % 2 == getBit(s, curr)) {
    ret += solve(curr + 1, (c + 1 + (getBit(x, curr) ^ 1)) >= 2, 1);
  }
  if ((c + 0 + (getBit(x, curr) ^ 0)) % 2 == getBit(s, curr)) {
    ret += solve(curr + 1, (c + 0 + (getBit(x, curr) ^ 0)) >= 2, f);
  }
  return ret;
}
int main(void) {
  cin >> s >> x;
  for (int i = 59; i >= 0; --i) {
    if (getBit(x, i) && !getBit(s, i)) {
      cout << 0 << "\n";
      return 0;
    } else if (getBit(s, i)) {
      n = i;
      break;
    }
  }
  memset(dp, -1, sizeof dp);
  long long ret = solve(0, 0, 0);
  if (x == s)
    cout << ret - 1 << "\n";
  else
    cout << ret << "\n";
  return 0;
}

```