#include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
const int logn = 20;
const int inf = (int)1e9;
const long long mod = (int)1e9 + 7;
const long long l_inf = (long long)4e18;
const long double pi = acosl(-1.0);
const long double eps = 1e-9;
bool check(long long s, long long x) {
  for (long long i = 50; i >= 0; i--) {
    if (x & (1ll << i)) continue;
    if (x + 2ll * (1ll << i) <= s) x += 2ll * (1ll << i);
  }
  return x == s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(15);
  cout << fixed;
  srand(566);
  long long s, x;
  cin >> s >> x;
  if (x > s || !check(s, x)) {
    cout << 0;
    exit(0);
  }
  long long cnt = 0;
  for (long long i = 0; (1ll << i) <= x; i++)
    if (x & (1ll << i)) cnt++;
  long long ans = 1ll << cnt;
  if (s == x) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 int maxn = 300005;
const int logn = 20;
const int inf = (int)1e9;
const long long mod = (int)1e9 + 7;
const long long l_inf = (long long)4e18;
const long double pi = acosl(-1.0);
const long double eps = 1e-9;
bool check(long long s, long long x) {
  for (long long i = 50; i >= 0; i--) {
    if (x & (1ll << i)) continue;
    if (x + 2ll * (1ll << i) <= s) x += 2ll * (1ll << i);
  }
  return x == s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(15);
  cout << fixed;
  srand(566);
  long long s, x;
  cin >> s >> x;
  if (x > s || !check(s, x)) {
    cout << 0;
    exit(0);
  }
  long long cnt = 0;
  for (long long i = 0; (1ll << i) <= x; i++)
    if (x & (1ll << i)) cnt++;
  long long ans = 1ll << cnt;
  if (s == x) ans -= 2;
  cout << ans;
  return 0;
}

```