#include <bits/stdc++.h>
using namespace std;
double esp = 1e-11;
const double PI = acos(-1.0);
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long long int MOD = (1ll << 63);
const int maxn = 1000010;
int main() {
  long long int s, x, ans;
  bitset<65> a, b, c;
  scanf("%I64d%I64d", &s, &x);
  if (s < x || ((s - x) & 1)) {
    printf("%d\n", 0);
    return 0;
  }
  long long int y = (s - x) >> 1;
  ans = 0;
  a = x;
  b = y;
  c = (~b) & a;
  if (c != a)
    ans = 0;
  else
    ans = 1ll << c.count();
  if (y == 0 && s == x) ans -= 2;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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;
double esp = 1e-11;
const double PI = acos(-1.0);
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long long int MOD = (1ll << 63);
const int maxn = 1000010;
int main() {
  long long int s, x, ans;
  bitset<65> a, b, c;
  scanf("%I64d%I64d", &s, &x);
  if (s < x || ((s - x) & 1)) {
    printf("%d\n", 0);
    return 0;
  }
  long long int y = (s - x) >> 1;
  ans = 0;
  a = x;
  b = y;
  c = (~b) & a;
  if (c != a)
    ans = 0;
  else
    ans = 1ll << c.count();
  if (y == 0 && s == x) ans -= 2;
  printf("%I64d\n", ans);
  return 0;
}

```