#include <bits/stdc++.h>
using namespace std;
long long s, x;
int digitxor[1050], digitand[1050], xorcnt, andcnt;
long long dfs(int cur) {
  if (cur == 0) return 1;
  if (digitxor[cur] == 1 && digitand[cur] == 1) return 0;
  if (digitxor[cur] == 0 && digitand[cur] == 1) return dfs(cur - 1);
  if (digitxor[cur] == 1 && digitand[cur] == 0) return dfs(cur - 1) * 2;
  if (digitxor[cur] == 0 && digitand[cur] == 0) return dfs(cur - 1);
}
int main() {
  scanf("%lld%lld", &s, &x);
  if ((s - x) % 2 == 1) printf("0\n");
  if (s < x) printf("0\n");
  if (s >= x && (s - x) % 2 == 0) {
    long long andsum = s - x;
    andsum /= 2;
    while (x) digitxor[++xorcnt] = x % 2, x /= 2;
    int flg = 2;
    while (andsum) {
      digitand[++andcnt] = andsum % 2, andsum /= 2;
      if (digitand[andcnt] == 1) flg = 0;
    }
    printf("%lld\n", dfs(max(xorcnt, andcnt)) - flg);
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 digitxor[1050], digitand[1050], xorcnt, andcnt;
long long dfs(int cur) {
  if (cur == 0) return 1;
  if (digitxor[cur] == 1 && digitand[cur] == 1) return 0;
  if (digitxor[cur] == 0 && digitand[cur] == 1) return dfs(cur - 1);
  if (digitxor[cur] == 1 && digitand[cur] == 0) return dfs(cur - 1) * 2;
  if (digitxor[cur] == 0 && digitand[cur] == 0) return dfs(cur - 1);
}
int main() {
  scanf("%lld%lld", &s, &x);
  if ((s - x) % 2 == 1) printf("0\n");
  if (s < x) printf("0\n");
  if (s >= x && (s - x) % 2 == 0) {
    long long andsum = s - x;
    andsum /= 2;
    while (x) digitxor[++xorcnt] = x % 2, x /= 2;
    int flg = 2;
    while (andsum) {
      digitand[++andcnt] = andsum % 2, andsum /= 2;
      if (digitand[andcnt] == 1) flg = 0;
    }
    printf("%lld\n", dfs(max(xorcnt, andcnt)) - flg);
  }
}

```