#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr static int mod = 1e9 + 7;
constexpr static int inf = (1 << 30) - 1;
constexpr static ll infll = (1LL << 61) - 1;
int program_init = (ios_base::sync_with_stdio(false), cin.tie(nullptr),
                    cout << fixed << setprecision(15), 0);
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
int main() {
  ll s, x;
  cin >> s >> x;
  ll K = (s - x) / 2;
  if (K < 0 || s != x + 2 * K || (x & K) != 0) {
    cout << 0 << "\n";
    return 0;
  }
  ll ans = 0;
  for (ll bit = 1; bit <= s; bit <<= 1) {
    if (!(K & bit) && (x & bit)) ans++;
  }
  cout << (1LL << ans) - (K == 0 ? 2 : 0) << "\n";
}

### Prompt

Generate 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;
using ll = long long;
constexpr static int mod = 1e9 + 7;
constexpr static int inf = (1 << 30) - 1;
constexpr static ll infll = (1LL << 61) - 1;
int program_init = (ios_base::sync_with_stdio(false), cin.tie(nullptr),
                    cout << fixed << setprecision(15), 0);
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
int main() {
  ll s, x;
  cin >> s >> x;
  ll K = (s - x) / 2;
  if (K < 0 || s != x + 2 * K || (x & K) != 0) {
    cout << 0 << "\n";
    return 0;
  }
  ll ans = 0;
  for (ll bit = 1; bit <= s; bit <<= 1) {
    if (!(K & bit) && (x & bit)) ans++;
  }
  cout << (1LL << ans) - (K == 0 ? 2 : 0) << "\n";
}

```