#include <bits/stdc++.h>
using namespace std;
vector<int> bs;
vector<int> bx;
long long f[60][2];
int main() {
  long long s, x;
  cin >> s >> x;
  long long ss = s;
  long long xx = x;
  if (x == 0) {
    if (s % 2 == 0) {
      cout << 1 << endl;
    } else {
      cout << 0 << endl;
    }
    return 0;
  }
  do {
    bs.push_back(s % 2);
    s /= 2;
  } while (s != 0);
  do {
    bx.push_back(x % 2);
    x /= 2;
  } while (x != 0);
  f[0][0] = 1;
  for (int i = 1; i <= bs.size(); ++i) {
    int cur_bit_s = (i <= bs.size() ? bs[i - 1] : 0);
    int cur_bit_x = (i <= bx.size() ? bx[i - 1] : 0);
    if (cur_bit_x == 0 && cur_bit_s == 0) {
      f[i][1] = f[i - 1][0];
      f[i][0] = f[i - 1][0];
    }
    if (cur_bit_x == 1 && cur_bit_s == 1) {
      f[i][1] = 0;
      f[i][0] = f[i - 1][0] * 2;
    }
    if (cur_bit_x == 0 && cur_bit_s == 1) {
      f[i][0] = f[i - 1][1];
      f[i][1] = f[i - 1][1];
    }
    if (cur_bit_x == 1 && cur_bit_s == 0) {
      f[i][1] = f[i - 1][1] * 2;
      f[i][0] = 0;
    }
  }
  if (bs.size() > bx.size()) {
    cout << f[bs.size() - 1][1] << endl;
  } else {
    if (xx == ss) {
      cout << max(f[bx.size()][0] - 2, 0ll) << endl;
    } else {
      cout << f[bx.size()][0] << endl;
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
vector<int> bs;
vector<int> bx;
long long f[60][2];
int main() {
  long long s, x;
  cin >> s >> x;
  long long ss = s;
  long long xx = x;
  if (x == 0) {
    if (s % 2 == 0) {
      cout << 1 << endl;
    } else {
      cout << 0 << endl;
    }
    return 0;
  }
  do {
    bs.push_back(s % 2);
    s /= 2;
  } while (s != 0);
  do {
    bx.push_back(x % 2);
    x /= 2;
  } while (x != 0);
  f[0][0] = 1;
  for (int i = 1; i <= bs.size(); ++i) {
    int cur_bit_s = (i <= bs.size() ? bs[i - 1] : 0);
    int cur_bit_x = (i <= bx.size() ? bx[i - 1] : 0);
    if (cur_bit_x == 0 && cur_bit_s == 0) {
      f[i][1] = f[i - 1][0];
      f[i][0] = f[i - 1][0];
    }
    if (cur_bit_x == 1 && cur_bit_s == 1) {
      f[i][1] = 0;
      f[i][0] = f[i - 1][0] * 2;
    }
    if (cur_bit_x == 0 && cur_bit_s == 1) {
      f[i][0] = f[i - 1][1];
      f[i][1] = f[i - 1][1];
    }
    if (cur_bit_x == 1 && cur_bit_s == 0) {
      f[i][1] = f[i - 1][1] * 2;
      f[i][0] = 0;
    }
  }
  if (bs.size() > bx.size()) {
    cout << f[bs.size() - 1][1] << endl;
  } else {
    if (xx == ss) {
      cout << max(f[bx.size()][0] - 2, 0ll) << endl;
    } else {
      cout << f[bx.size()][0] << endl;
    }
  }
  return 0;
}

```