#include <bits/stdc++.h>
using namespace std;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
using ll = long long;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
ll solve(ll s, ll x) {
  if (s < x) {
    return 0;
  }
  ll d = s - x;
  if (d % 2) return 0;
  d /= 2;
  if (x & d) return 0;
  ll res = 1LL << __builtin_popcountll(x);
  if (d == 0) res -= 2;
  return max(res, 0LL);
}
ll brute(ll s, ll x) {
  int ans = 0;
  for (int a = 1; a <= (int)(s - 1); a++) {
    int b = s - a;
    if ((a ^ b) == x) ans++;
  }
  return ans;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  if (false) {
    for (int i = 0; i < (int)(100); i++)
      for (int j = 0; j < (int)(100); j++) {
        assert(solve(i, j) == brute(i, j));
      }
  }
  ll s, x;
  cin >> s >> x;
  cout << solve(s, x) << endl;
  true;
}

### Prompt

In Cpp, your task is to solve 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 <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
using ll = long long;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
ll solve(ll s, ll x) {
  if (s < x) {
    return 0;
  }
  ll d = s - x;
  if (d % 2) return 0;
  d /= 2;
  if (x & d) return 0;
  ll res = 1LL << __builtin_popcountll(x);
  if (d == 0) res -= 2;
  return max(res, 0LL);
}
ll brute(ll s, ll x) {
  int ans = 0;
  for (int a = 1; a <= (int)(s - 1); a++) {
    int b = s - a;
    if ((a ^ b) == x) ans++;
  }
  return ans;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  if (false) {
    for (int i = 0; i < (int)(100); i++)
      for (int j = 0; j < (int)(100); j++) {
        assert(solve(i, j) == brute(i, j));
      }
  }
  ll s, x;
  cin >> s >> x;
  cout << solve(s, x) << endl;
  true;
}

```