#include <bits/stdc++.h>
using namespace std;
long long pw2[64];
int main() {
  pw2[0] = 1;
  for (int i = 1; i < 64; i++) pw2[i] = pw2[i - 1] * 2;
  long long s, x;
  cin >> s >> x;
  vector<int> bits;
  bits.clear();
  long long tmpx = x;
  while (tmpx > 0) {
    bits.push_back(tmpx % 2);
    tmpx /= 2;
  }
  long long prod = 1, sumover = 0;
  for (int i = 0; i < int(bits.size()); i++) {
    if (bits[i] == 1) {
      sumover += pw2[i];
      prod *= 2;
    }
  }
  if (sumover > s) {
    cout << 0 << endl;
    return 0;
  }
  if (sumover == s) {
    cout << prod - 2 << endl;
    return 0;
  }
  s -= sumover;
  if (s & 1) {
    cout << 0 << endl;
    return 0;
  }
  s /= 2;
  tmpx = s;
  vector<int> getbits;
  getbits.clear();
  while (tmpx > 0) {
    getbits.push_back(tmpx % 2);
    tmpx /= 2;
  }
  for (int i = 0; i < int(getbits.size()); i++) {
    if (getbits[i] == 1 && (i < int(bits.size()) && bits[i] == 1)) {
      cout << 0 << endl;
      return 0;
    }
  }
  cout << prod << endl;
}

### Prompt

Please create a solution in CPP 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 pw2[64];
int main() {
  pw2[0] = 1;
  for (int i = 1; i < 64; i++) pw2[i] = pw2[i - 1] * 2;
  long long s, x;
  cin >> s >> x;
  vector<int> bits;
  bits.clear();
  long long tmpx = x;
  while (tmpx > 0) {
    bits.push_back(tmpx % 2);
    tmpx /= 2;
  }
  long long prod = 1, sumover = 0;
  for (int i = 0; i < int(bits.size()); i++) {
    if (bits[i] == 1) {
      sumover += pw2[i];
      prod *= 2;
    }
  }
  if (sumover > s) {
    cout << 0 << endl;
    return 0;
  }
  if (sumover == s) {
    cout << prod - 2 << endl;
    return 0;
  }
  s -= sumover;
  if (s & 1) {
    cout << 0 << endl;
    return 0;
  }
  s /= 2;
  tmpx = s;
  vector<int> getbits;
  getbits.clear();
  while (tmpx > 0) {
    getbits.push_back(tmpx % 2);
    tmpx /= 2;
  }
  for (int i = 0; i < int(getbits.size()); i++) {
    if (getbits[i] == 1 && (i < int(bits.size()) && bits[i] == 1)) {
      cout << 0 << endl;
      return 0;
    }
  }
  cout << prod << endl;
}

```