#include <bits/stdc++.h>
using namespace std;
const int B = 40;
long long int dp[B + 5][2][4];
int main() {
  long long S, X;
  scanf("%lld%lld", &S, &X);
  dp[0][0][0] = 1;
  for (int i = 0; i < B; i++) {
    for (int bor = 0; bor < 2; bor++) {
      for (int bit = 0; bit < 4; bit++) {
        for (int x = 0; x < 2; x++) {
          for (int y = 0; y < 2; y++) {
            int nbor = (bor + x + y) / 2;
            int ad = (bor + x + y) % 2;
            int xo = (x ^ y);
            int nbit = bit;
            if (x == 1) nbit |= 1;
            if (y == 1) nbit |= 2;
            if ((S >> i & 1) != ad) continue;
            if ((X >> i & 1) != xo) continue;
            dp[i + 1][nbor][nbit] += dp[i][bor][bit];
          }
        }
      }
    }
  }
  long long ans = dp[B][0][3];
  printf("%lld\n", ans);
  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>
using namespace std;
const int B = 40;
long long int dp[B + 5][2][4];
int main() {
  long long S, X;
  scanf("%lld%lld", &S, &X);
  dp[0][0][0] = 1;
  for (int i = 0; i < B; i++) {
    for (int bor = 0; bor < 2; bor++) {
      for (int bit = 0; bit < 4; bit++) {
        for (int x = 0; x < 2; x++) {
          for (int y = 0; y < 2; y++) {
            int nbor = (bor + x + y) / 2;
            int ad = (bor + x + y) % 2;
            int xo = (x ^ y);
            int nbit = bit;
            if (x == 1) nbit |= 1;
            if (y == 1) nbit |= 2;
            if ((S >> i & 1) != ad) continue;
            if ((X >> i & 1) != xo) continue;
            dp[i + 1][nbor][nbit] += dp[i][bor][bit];
          }
        }
      }
    }
  }
  long long ans = dp[B][0][3];
  printf("%lld\n", ans);
  return 0;
}

```