#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int num[101010];
long long int ans = 1, chousei = 0;
int main() {
  long long int sm, xr, hoge;
  bool flag = true;
  cin >> sm >> xr;
  hoge = sm - xr;
  if (hoge < 0) flag = false;
  vector<int> bits1, bits2;
  while (hoge) {
    int x = (hoge % 2);
    hoge /= 2;
    bits1.push_back(x);
  }
  if (sm == xr) chousei = -2;
  int cnt = 0;
  bits2.push_back(1);
  while (xr) {
    int x = (xr % 2);
    xr /= 2;
    bits2.push_back(x);
    if (x == 1) cnt++;
  }
  for (int i = 0; i < min(bits1.size(), bits2.size()); i++) {
    if (bits1[i] == 1 && bits2[i] == 1) {
      flag = false;
    }
  }
  if (flag) {
    for (int i = 0; i < cnt; i++) {
      ans *= 2;
    }
    cout << ans + chousei << endl;
  } else
    cout << "0" << endl;
}

### Prompt

Your task is to create 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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int num[101010];
long long int ans = 1, chousei = 0;
int main() {
  long long int sm, xr, hoge;
  bool flag = true;
  cin >> sm >> xr;
  hoge = sm - xr;
  if (hoge < 0) flag = false;
  vector<int> bits1, bits2;
  while (hoge) {
    int x = (hoge % 2);
    hoge /= 2;
    bits1.push_back(x);
  }
  if (sm == xr) chousei = -2;
  int cnt = 0;
  bits2.push_back(1);
  while (xr) {
    int x = (xr % 2);
    xr /= 2;
    bits2.push_back(x);
    if (x == 1) cnt++;
  }
  for (int i = 0; i < min(bits1.size(), bits2.size()); i++) {
    if (bits1[i] == 1 && bits2[i] == 1) {
      flag = false;
    }
  }
  if (flag) {
    for (int i = 0; i < cnt; i++) {
      ans *= 2;
    }
    cout << ans + chousei << endl;
  } else
    cout << "0" << endl;
}

```