#include <bits/stdc++.h>
using namespace std;
const int NUM = 50;
const int mod = 1e9 + 7;
long long INF = 1e9 + 50;
pair<int, int> pinf = {INF, INF};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long s, x, t;
  cin >> s >> x;
  t = s;
  long long bits[NUM];
  long long k = 0;
  for (int i = 0; i < NUM; i++) {
    bits[i] = (x >> i) % 2;
    if (bits[i] == 1) {
      k++;
      s -= ((long long)1 << i);
    }
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  long long ans = (long long)1 << k;
  for (int i = 0; i < NUM; i++) {
    if (bits[i] == 0 && (i == 0 || bits[i - 1] == 1)) {
      if ((s >> i) % 2 == 1) {
        cout << 0;
        return 0;
      }
    }
    if (bits[i] == 1 && (i == 0 || bits[i - 1] == 1)) {
      if ((s >> i) % 2 == 1) {
        cout << 0;
        return 0;
      }
    }
  }
  if (t == x) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 NUM = 50;
const int mod = 1e9 + 7;
long long INF = 1e9 + 50;
pair<int, int> pinf = {INF, INF};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long s, x, t;
  cin >> s >> x;
  t = s;
  long long bits[NUM];
  long long k = 0;
  for (int i = 0; i < NUM; i++) {
    bits[i] = (x >> i) % 2;
    if (bits[i] == 1) {
      k++;
      s -= ((long long)1 << i);
    }
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  long long ans = (long long)1 << k;
  for (int i = 0; i < NUM; i++) {
    if (bits[i] == 0 && (i == 0 || bits[i - 1] == 1)) {
      if ((s >> i) % 2 == 1) {
        cout << 0;
        return 0;
      }
    }
    if (bits[i] == 1 && (i == 0 || bits[i - 1] == 1)) {
      if ((s >> i) % 2 == 1) {
        cout << 0;
        return 0;
      }
    }
  }
  if (t == x) ans -= 2;
  cout << ans;
  return 0;
}

```