#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 233;
struct poi {
  int x, y;
} a[maxn];
int i, j, k, n, m;
int ra, fh;
char rx;
inline int read() {
  rx = getchar(), ra = 0, fh = 1;
  while ((rx < '0' || rx > '9') && rx != '-') rx = getchar();
  if (rx == '-') fh = -1, rx = getchar();
  while (rx >= '0' && rx <= '9') ra = ra * 10 + rx - 48, rx = getchar();
  return ra * fh;
}
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  if (x > s) return puts("0"), 0;
  long long rest = s - x;
  if ((rest & 1) || ((rest >> 1) & x)) return puts("0"), 0;
  for (i = 0; i < 41; i++)
    if (x >> i & 1) k++;
  printf("%I64d\n", (1ll << k) - (!rest ? 2 : 0));
}

### 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 maxn = 2e5 + 233;
struct poi {
  int x, y;
} a[maxn];
int i, j, k, n, m;
int ra, fh;
char rx;
inline int read() {
  rx = getchar(), ra = 0, fh = 1;
  while ((rx < '0' || rx > '9') && rx != '-') rx = getchar();
  if (rx == '-') fh = -1, rx = getchar();
  while (rx >= '0' && rx <= '9') ra = ra * 10 + rx - 48, rx = getchar();
  return ra * fh;
}
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  if (x > s) return puts("0"), 0;
  long long rest = s - x;
  if ((rest & 1) || ((rest >> 1) & x)) return puts("0"), 0;
  for (i = 0; i < 41; i++)
    if (x >> i & 1) k++;
  printf("%I64d\n", (1ll << k) - (!rest ? 2 : 0));
}

```