#include <bits/stdc++.h>
using namespace std;
const long long linf = 1e18 + 5;
const int mod = (int)1e9 + 7;
const int logN = 17;
const int inf = 1e9;
const int N = 2e5 + 5;
int main() {
  vector<long long> tt;
  long long ans = 1, x, y;
  scanf("%lld %lld", &x, &y);
  for (int i = 60; i >= 0; i--) {
    long long t = 1ll << (long long)(i);
    if (t & y) {
      ans *= 2;
      x -= t;
    } else {
      tt.push_back(t * 2LL);
    }
  }
  if (x == 0) ans -= 2;
  for (__typeof(tt.begin()) it = tt.begin(); it != tt.end(); it++) {
    if (*it <= x) {
      x -= *it;
    }
  }
  if (!x)
    cout << ans << '\n';
  else
    cout << 0 << '\n';
  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;
const long long linf = 1e18 + 5;
const int mod = (int)1e9 + 7;
const int logN = 17;
const int inf = 1e9;
const int N = 2e5 + 5;
int main() {
  vector<long long> tt;
  long long ans = 1, x, y;
  scanf("%lld %lld", &x, &y);
  for (int i = 60; i >= 0; i--) {
    long long t = 1ll << (long long)(i);
    if (t & y) {
      ans *= 2;
      x -= t;
    } else {
      tt.push_back(t * 2LL);
    }
  }
  if (x == 0) ans -= 2;
  for (__typeof(tt.begin()) it = tt.begin(); it != tt.end(); it++) {
    if (*it <= x) {
      x -= *it;
    }
  }
  if (!x)
    cout << ans << '\n';
  else
    cout << 0 << '\n';
  return 0;
}

```