#include <bits/stdc++.h>
using namespace std;
int N;
long long S, X, dp[60][2][2][2];
long long solve(int p, int ta, int tb, int fl) {
  if (p >= N) return (!fl && ta && tb);
  long long &res = dp[p][ta][tb][fl];
  if (res != -1) return res;
  res = 0;
  int x = (X & (1LL << p)) > 0LL;
  int s = (S & (1LL << p)) > 0LL;
  for (int i = 0; i < (int)2; i++)
    for (int j = 0; j < (int)2; j++) {
      int b = i + j;
      if ((b & 1) != x) continue;
      if (((b + fl) & 1) != s) continue;
      int nfl = (b + fl >= 2);
      res += solve(p + 1, ta | i, tb | j, nfl);
    }
  return res;
}
int main() {
  scanf("%I64d%I64d", &S, &X);
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < (int)60; i++)
    if ((1LL << i) > max(S, X)) {
      N = i;
      break;
    }
  printf("%I64d\n", solve(0, 0, 0, 0));
}

### 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 N;
long long S, X, dp[60][2][2][2];
long long solve(int p, int ta, int tb, int fl) {
  if (p >= N) return (!fl && ta && tb);
  long long &res = dp[p][ta][tb][fl];
  if (res != -1) return res;
  res = 0;
  int x = (X & (1LL << p)) > 0LL;
  int s = (S & (1LL << p)) > 0LL;
  for (int i = 0; i < (int)2; i++)
    for (int j = 0; j < (int)2; j++) {
      int b = i + j;
      if ((b & 1) != x) continue;
      if (((b + fl) & 1) != s) continue;
      int nfl = (b + fl >= 2);
      res += solve(p + 1, ta | i, tb | j, nfl);
    }
  return res;
}
int main() {
  scanf("%I64d%I64d", &S, &X);
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < (int)60; i++)
    if ((1LL << i) > max(S, X)) {
      N = i;
      break;
    }
  printf("%I64d\n", solve(0, 0, 0, 0));
}

```