#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const double EPS = 1e-7;
const double PI = acos(-1.0);
const long long INF = (long long)2e9 + 1;
const long long LINF = (long long)8e18;
const long long MM = (long long)1e9 + 7;
int solve();
void gen();
int main() {
  solve();
  return 0;
}
const int MAXN = 1e5 + 7;
const int MAXC = 5e7;
int solve() {
  long long a, b;
  cin >> a >> b;
  if ((a - b) % 2 || a < b) {
    cout << 0;
    return 0;
  }
  long long cur = abs(a - b) / 2;
  long long cnt = 0;
  for (long long i = 0; i < 60; i++) {
    if ((1LL << i) & b) {
      cnt++;
    }
  }
  for (long long i = 60; i >= 0; i--) {
    if (!(b & (1LL << i)) && cur & (1LL << i)) cur -= (1LL << i);
  }
  cur ? cout << 0 : cout << (a - b != 0 ? (1LL << cnt) : (1LL << cnt) - 2);
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const double EPS = 1e-7;
const double PI = acos(-1.0);
const long long INF = (long long)2e9 + 1;
const long long LINF = (long long)8e18;
const long long MM = (long long)1e9 + 7;
int solve();
void gen();
int main() {
  solve();
  return 0;
}
const int MAXN = 1e5 + 7;
const int MAXC = 5e7;
int solve() {
  long long a, b;
  cin >> a >> b;
  if ((a - b) % 2 || a < b) {
    cout << 0;
    return 0;
  }
  long long cur = abs(a - b) / 2;
  long long cnt = 0;
  for (long long i = 0; i < 60; i++) {
    if ((1LL << i) & b) {
      cnt++;
    }
  }
  for (long long i = 60; i >= 0; i--) {
    if (!(b & (1LL << i)) && cur & (1LL << i)) cur -= (1LL << i);
  }
  cur ? cout << 0 : cout << (a - b != 0 ? (1LL << cnt) : (1LL << cnt) - 2);
  return 0;
}

```