#include <bits/stdc++.h>
using namespace std;
const long long MAXBIT = 62ll;
vector<long long> toBit(long long x) {
  vector<long long> ans(MAXBIT);
  for (long long i = 0; i < MAXBIT; ++i) {
    ans[i] = (x & 1ll);
    x >>= 1;
  }
  return ans;
}
int main() {
  long long s, x;
  cin >> s >> x;
  vector<long long> xr = toBit(x);
  long long cur_sum = 0;
  long long cnt = 0;
  for (long long i = 0; i < MAXBIT; ++i) {
    cur_sum += xr[i] * (1ll << i);
    cnt += xr[i];
  }
  if (cur_sum > s) {
    cout << 0 << endl;
    return 0;
  }
  long long need = s - cur_sum;
  if (need % 2ll == 1ll) {
    cout << 0 << endl;
    return 0;
  }
  need >>= 1ll;
  vector<long long> nbit = toBit(need);
  for (long long i = 0; i < MAXBIT; ++i) {
    if (nbit[i] == 1ll && xr[i] == 1ll) {
      cout << 0 << endl;
      return 0;
    }
  }
  long long ans = (1ll << cnt);
  if (need == 0) ans -= 2;
  cout << ans << endl;
  return 0;
  return 0;
}

### Prompt

Develop a solution in CPP 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 MAXBIT = 62ll;
vector<long long> toBit(long long x) {
  vector<long long> ans(MAXBIT);
  for (long long i = 0; i < MAXBIT; ++i) {
    ans[i] = (x & 1ll);
    x >>= 1;
  }
  return ans;
}
int main() {
  long long s, x;
  cin >> s >> x;
  vector<long long> xr = toBit(x);
  long long cur_sum = 0;
  long long cnt = 0;
  for (long long i = 0; i < MAXBIT; ++i) {
    cur_sum += xr[i] * (1ll << i);
    cnt += xr[i];
  }
  if (cur_sum > s) {
    cout << 0 << endl;
    return 0;
  }
  long long need = s - cur_sum;
  if (need % 2ll == 1ll) {
    cout << 0 << endl;
    return 0;
  }
  need >>= 1ll;
  vector<long long> nbit = toBit(need);
  for (long long i = 0; i < MAXBIT; ++i) {
    if (nbit[i] == 1ll && xr[i] == 1ll) {
      cout << 0 << endl;
      return 0;
    }
  }
  long long ans = (1ll << cnt);
  if (need == 0) ans -= 2;
  cout << ans << endl;
  return 0;
  return 0;
}

```