#include <bits/stdc++.h>
using namespace std;
struct $ {
  $() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
} $;
const int Maxn = 200005;
int f(int a, int b) {
  int ret = 0;
  for (int i = 1; i < 100; i++) {
    int x = i;
    int y = a - i;
    if (y > 0) {
      if ((y ^ x) == b) {
        ret++;
      }
    }
  }
  return ret;
}
int main() {
  long long s, x, Ans = 0, t;
  cin >> s >> x;
  for (int i = 0; i < 60; i++) {
    if ((x & (1LL << i))) Ans++;
  }
  t = (s - x);
  if (t & 1LL) {
    cout << 0 << endl;
    return 0;
  }
  t = t >> 1LL;
  if (t == 0) {
    long long Res1 = (1LL << Ans);
    cout << (Res1 - 2) << endl;
    return 0;
  }
  bool Ok = true;
  for (long long i = 0; i < 60; i++) {
    if ((t & (1LL << i)) & (x)) Ok = false;
  }
  if (!Ok) {
    cout << 0 << endl;
    return 0;
  } else {
    long long Res;
    Res = (1LL << Ans);
    cout << Res << endl;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
struct $ {
  $() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
} $;
const int Maxn = 200005;
int f(int a, int b) {
  int ret = 0;
  for (int i = 1; i < 100; i++) {
    int x = i;
    int y = a - i;
    if (y > 0) {
      if ((y ^ x) == b) {
        ret++;
      }
    }
  }
  return ret;
}
int main() {
  long long s, x, Ans = 0, t;
  cin >> s >> x;
  for (int i = 0; i < 60; i++) {
    if ((x & (1LL << i))) Ans++;
  }
  t = (s - x);
  if (t & 1LL) {
    cout << 0 << endl;
    return 0;
  }
  t = t >> 1LL;
  if (t == 0) {
    long long Res1 = (1LL << Ans);
    cout << (Res1 - 2) << endl;
    return 0;
  }
  bool Ok = true;
  for (long long i = 0; i < 60; i++) {
    if ((t & (1LL << i)) & (x)) Ok = false;
  }
  if (!Ok) {
    cout << 0 << endl;
    return 0;
  } else {
    long long Res;
    Res = (1LL << Ans);
    cout << Res << endl;
  }
  return 0;
}

```