#include <bits/stdc++.h>
long long s, x, d, dp[51ll][2];
int main() {
  std::cin >> s >> x;
  if ((s - x) % 2 == 1ll || s < x) return puts("0"), 0;
  d = s - x >> 1ll;
  if ((d & (~x)) != d) return puts("0"), 0;
  dp[40][1] = 1;
  for (int i = 41; i > 0; i--) {
    for (int j = 0; j < 2; j++) {
      if (!dp[i][j]) continue;
      int now = (d >> i - 1ll) & 1ll, limit = (s >> i - 1ll) & 1ll;
      for (int k = now; k < 2; k++) {
        if (j && (k > limit)) continue;
        if (now == 0 && (k && ((~x >> i - 1ll) & 1ll))) continue;
        dp[i - 1][j && (k == limit)] += dp[i][j];
      }
    }
  }
  std::cout << dp[0][0] - (x == s) << std::endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
long long s, x, d, dp[51ll][2];
int main() {
  std::cin >> s >> x;
  if ((s - x) % 2 == 1ll || s < x) return puts("0"), 0;
  d = s - x >> 1ll;
  if ((d & (~x)) != d) return puts("0"), 0;
  dp[40][1] = 1;
  for (int i = 41; i > 0; i--) {
    for (int j = 0; j < 2; j++) {
      if (!dp[i][j]) continue;
      int now = (d >> i - 1ll) & 1ll, limit = (s >> i - 1ll) & 1ll;
      for (int k = now; k < 2; k++) {
        if (j && (k > limit)) continue;
        if (now == 0 && (k && ((~x >> i - 1ll) & 1ll))) continue;
        dp[i - 1][j && (k == limit)] += dp[i][j];
      }
    }
  }
  std::cout << dp[0][0] - (x == s) << std::endl;
  return 0;
}

```