#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int inf = 1e9 + 5;
long long powmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 2e5 + 5;
long long s, x;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s >> x;
  if ((s - x) & 1) {
    cout << "0\n";
    return 0;
  }
  long long y = (s - x) / 2;
  if (y & x) {
    cout << "0\n";
    return 0;
  }
  long long ways = 0;
  for (int i = 0; i <= 40; i++) {
    long long setbit = pow(2 * 1LL, i * 1LL);
    if (x & setbit) ways++;
  }
  ways = pow(2 * 1LL, ways);
  if (y == 0) ways -= 2;
  cout << ways << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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 mod = 1e9 + 7;
const int inf = 1e9 + 5;
long long powmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 2e5 + 5;
long long s, x;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s >> x;
  if ((s - x) & 1) {
    cout << "0\n";
    return 0;
  }
  long long y = (s - x) / 2;
  if (y & x) {
    cout << "0\n";
    return 0;
  }
  long long ways = 0;
  for (int i = 0; i <= 40; i++) {
    long long setbit = pow(2 * 1LL, i * 1LL);
    if (x & setbit) ways++;
  }
  ways = pow(2 * 1LL, ways);
  if (y == 0) ways -= 2;
  cout << ways << "\n";
  return 0;
}

```