#include <bits/stdc++.h>
using namespace std;
long long dp[50][2];
int main() {
  long long s, x;
  scanf("%lld %lld", &s, &x);
  long long menos = 0;
  if ((s ^ 0) == x) menos += 2;
  dp[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    if ((s & 1) == 0 && (x & 1) == 1) dp[i][1] += dp[i - 1][1] * 2;
    if ((s & 1) == 1 && (x & 1) == 0) {
      dp[i][0] += dp[i - 1][1];
      dp[i][1] += dp[i - 1][1];
    }
    if ((s & 1) == 1 && (x & 1) == 1) dp[i][0] += 2 * dp[i - 1][0];
    if ((s & 1) == 0 && (x & 1) == 0) {
      dp[i][0] += dp[i - 1][0];
      dp[i][1] += dp[i - 1][0];
    }
    s = s >> 1;
    x = x >> 1;
  }
  printf("%lld", dp[49][0] - menos);
  return 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;
long long dp[50][2];
int main() {
  long long s, x;
  scanf("%lld %lld", &s, &x);
  long long menos = 0;
  if ((s ^ 0) == x) menos += 2;
  dp[0][0] = 1;
  for (int i = 1; i < 50; i++) {
    if ((s & 1) == 0 && (x & 1) == 1) dp[i][1] += dp[i - 1][1] * 2;
    if ((s & 1) == 1 && (x & 1) == 0) {
      dp[i][0] += dp[i - 1][1];
      dp[i][1] += dp[i - 1][1];
    }
    if ((s & 1) == 1 && (x & 1) == 1) dp[i][0] += 2 * dp[i - 1][0];
    if ((s & 1) == 0 && (x & 1) == 0) {
      dp[i][0] += dp[i - 1][0];
      dp[i][1] += dp[i - 1][0];
    }
    s = s >> 1;
    x = x >> 1;
  }
  printf("%lld", dp[49][0] - menos);
  return 0;
}

```