#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200100;
long long s, k;
long long bit[110];
int b[110];
int w1;
int main() {
  w1 = 0;
  bit[0] = 1;
  for (int i = 1; i <= 63; i++) bit[i] = bit[i - 1] * 2;
  memset(b, 0, sizeof(b));
  scanf("%I64d%I64d", &s, &k);
  for (int i = 0; i <= 63; i++) {
    if (bit[i] & k) {
      s -= bit[i];
      w1++;
    } else {
      b[i + 1] = 1;
    }
  }
  bool flag = true;
  for (int i = 0; i <= 63; i++) {
    if ((s & bit[i]) != 0) {
      if (!b[i]) {
        flag = false;
      }
    }
  }
  if (!flag) {
    printf("0\n");
  } else {
    long long ans = 1;
    for (int i = 1; i <= w1; i++) {
      ans *= 2;
    }
    if (!s) {
      ans -= 2;
    }
    printf("%I64d\n", ans);
  }
}

### Prompt

Please formulate 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;
const int MAX_N = 200100;
long long s, k;
long long bit[110];
int b[110];
int w1;
int main() {
  w1 = 0;
  bit[0] = 1;
  for (int i = 1; i <= 63; i++) bit[i] = bit[i - 1] * 2;
  memset(b, 0, sizeof(b));
  scanf("%I64d%I64d", &s, &k);
  for (int i = 0; i <= 63; i++) {
    if (bit[i] & k) {
      s -= bit[i];
      w1++;
    } else {
      b[i + 1] = 1;
    }
  }
  bool flag = true;
  for (int i = 0; i <= 63; i++) {
    if ((s & bit[i]) != 0) {
      if (!b[i]) {
        flag = false;
      }
    }
  }
  if (!flag) {
    printf("0\n");
  } else {
    long long ans = 1;
    for (int i = 1; i <= w1; i++) {
      ans *= 2;
    }
    if (!s) {
      ans -= 2;
    }
    printf("%I64d\n", ans);
  }
}

```