#include <bits/stdc++.h>
using namespace std;
int main(void) {
  long long s, x;
  while (scanf("%I64d %I64d", &s, &x) != EOF) {
    long long res = 1;
    int curs, curx, was = 0, k;
    for (k = 60; k >= 0; k--) {
      curx = (x >> k) & 1;
      curs = (s >> k) & 1;
      res <<= curx;
      if (was == 0) {
        if (curs == 0 && curx == 1) res = 0;
        if (curs == 1 && curx == 0) was = 1;
      } else {
        if (curs == 1 && curx == 1) res = 0;
        if (curs == 0 && curx == 0) was = 0;
      }
    }
    if (s == x) res -= 2;
    if (was) res = 0;
    printf("%I64d\n", res);
  }
  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;
int main(void) {
  long long s, x;
  while (scanf("%I64d %I64d", &s, &x) != EOF) {
    long long res = 1;
    int curs, curx, was = 0, k;
    for (k = 60; k >= 0; k--) {
      curx = (x >> k) & 1;
      curs = (s >> k) & 1;
      res <<= curx;
      if (was == 0) {
        if (curs == 0 && curx == 1) res = 0;
        if (curs == 1 && curx == 0) was = 1;
      } else {
        if (curs == 1 && curx == 1) res = 0;
        if (curs == 0 && curx == 0) was = 0;
      }
    }
    if (s == x) res -= 2;
    if (was) res = 0;
    printf("%I64d\n", res);
  }
  return 0;
}

```