#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  scanf("%lld%lld", &s, &x);
  long long s_high = s >> 20LL;
  long long s_low = s & ((1 << 20LL) - 1);
  long long x_high = x >> 20LL;
  long long x_low = x & ((1 << 20LL) - 1);
  long long no_carry = 0, carry = 0;
  for (long long a_high = 0; a_high < (1LL << 20LL); a_high++) {
    long long b_high = a_high ^ x_high;
    if (a_high + b_high == s_high) no_carry++;
    if (a_high + b_high + 1LL == s_high) carry++;
  }
  long long ans = 0;
  for (long long a_low = 0; a_low < (1LL << 20LL); a_low++) {
    long long b_low = a_low ^ x_low;
    if (a_low + b_low == s_low) ans += no_carry;
    if (a_low + b_low == s_low + (1LL << 20LL)) ans += carry;
  }
  if (s == x) ans -= 2LL;
  printf("%lld", ans);
  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 s, x;
int main() {
  scanf("%lld%lld", &s, &x);
  long long s_high = s >> 20LL;
  long long s_low = s & ((1 << 20LL) - 1);
  long long x_high = x >> 20LL;
  long long x_low = x & ((1 << 20LL) - 1);
  long long no_carry = 0, carry = 0;
  for (long long a_high = 0; a_high < (1LL << 20LL); a_high++) {
    long long b_high = a_high ^ x_high;
    if (a_high + b_high == s_high) no_carry++;
    if (a_high + b_high + 1LL == s_high) carry++;
  }
  long long ans = 0;
  for (long long a_low = 0; a_low < (1LL << 20LL); a_low++) {
    long long b_low = a_low ^ x_low;
    if (a_low + b_low == s_low) ans += no_carry;
    if (a_low + b_low == s_low + (1LL << 20LL)) ans += carry;
  }
  if (s == x) ans -= 2LL;
  printf("%lld", ans);
  return 0;
}

```