#include <bits/stdc++.h>
using namespace std;
int brute(int s, int x) {
  int ans = 0;
  for (int i = 1; i <= s; ++i)
    for (int j = 1; j <= s; ++j)
      if (i + j == s && (i ^ j) == x) ++ans;
  return ans;
}
long long ways0(long long s, long long x) {
  if (s == x) return 2;
  return 0;
}
long long solve(long long s, long long x) {
  long long dp[60][2] = {};
  if (s % 2 != x % 2) {
    return 0;
  }
  if (x & (1LL << 0)) {
    dp[0][0] = 2;
  } else {
    dp[0][0] = 1;
    dp[0][1] = 1;
  }
  for (int i = 1; i < 60; ++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][0] += dp[i - 1][1];
        dp[i][1] += dp[i - 1][1];
      } else {
        dp[i][0] += dp[i - 1][0];
        dp[i][1] += dp[i - 1][0];
      }
    }
  }
  return dp[59][0] - ways0(s, x);
}
int main() {
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x) << '\n';
}

### Prompt

Generate 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;
int brute(int s, int x) {
  int ans = 0;
  for (int i = 1; i <= s; ++i)
    for (int j = 1; j <= s; ++j)
      if (i + j == s && (i ^ j) == x) ++ans;
  return ans;
}
long long ways0(long long s, long long x) {
  if (s == x) return 2;
  return 0;
}
long long solve(long long s, long long x) {
  long long dp[60][2] = {};
  if (s % 2 != x % 2) {
    return 0;
  }
  if (x & (1LL << 0)) {
    dp[0][0] = 2;
  } else {
    dp[0][0] = 1;
    dp[0][1] = 1;
  }
  for (int i = 1; i < 60; ++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][0] += dp[i - 1][1];
        dp[i][1] += dp[i - 1][1];
      } else {
        dp[i][0] += dp[i - 1][0];
        dp[i][1] += dp[i - 1][0];
      }
    }
  }
  return dp[59][0] - ways0(s, x);
}
int main() {
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x) << '\n';
}

```