#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = (int)1e6 + 7;
const int MAXN = (int)2e6 + 7;
map<pair<long long, long long>, long long> used[1111];
long long calc(long long second, long long x, int bit) {
  if (bit == -1) {
    if (second == 0) return 1;
    return 0;
  }
  if (second < 0) return 0;
  long long cur = (x >> bit) & 1;
  long long in = (second >> bit) & 1;
  if (cur != in) return 0;
  if (cur == 1) {
    return 2ll * calc(second - (1ll << bit), x, bit - 1);
  } else {
    return calc(second - 2ll * (1ll << bit), x, bit - 1) +
           calc(second, x, bit - 1);
  }
}
long long slow(long long second, long long x) {
  long long lolo = 0;
  for (long long i = 1; i < second; i++) {
    if ((i ^ (second - i)) == x) {
      lolo++;
    }
  }
  return lolo;
}
int main() {
  long long second, x;
  cin >> second >> x;
  vector<int> a;
  vector<int> b;
  for (int i = 0; i < 59; i++) {
    a.push_back((second >> i) & 1);
    b.push_back((x >> i) & 1);
  }
  int same = 1;
  long long ans = 1;
  for (int i = 0; i < 59; i++) {
    if (a[i] != b[i]) {
      if (i == 0) {
        cout << 0;
        return 0;
      }
      if (a[i - 1] == b[i - 1] && a[i - 1]) {
        cout << 0;
        return 0;
      }
      if (a[i] == 0) {
        ans += ans;
      }
    } else {
      if (i != 0 && a[i - 1] != b[i - 1] && !a[i - 1]) {
        cout << 0;
        return 0;
      }
      if (a[i] == 1) {
        ans += ans;
      }
    }
  }
  if (second == x) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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;
const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = (int)1e6 + 7;
const int MAXN = (int)2e6 + 7;
map<pair<long long, long long>, long long> used[1111];
long long calc(long long second, long long x, int bit) {
  if (bit == -1) {
    if (second == 0) return 1;
    return 0;
  }
  if (second < 0) return 0;
  long long cur = (x >> bit) & 1;
  long long in = (second >> bit) & 1;
  if (cur != in) return 0;
  if (cur == 1) {
    return 2ll * calc(second - (1ll << bit), x, bit - 1);
  } else {
    return calc(second - 2ll * (1ll << bit), x, bit - 1) +
           calc(second, x, bit - 1);
  }
}
long long slow(long long second, long long x) {
  long long lolo = 0;
  for (long long i = 1; i < second; i++) {
    if ((i ^ (second - i)) == x) {
      lolo++;
    }
  }
  return lolo;
}
int main() {
  long long second, x;
  cin >> second >> x;
  vector<int> a;
  vector<int> b;
  for (int i = 0; i < 59; i++) {
    a.push_back((second >> i) & 1);
    b.push_back((x >> i) & 1);
  }
  int same = 1;
  long long ans = 1;
  for (int i = 0; i < 59; i++) {
    if (a[i] != b[i]) {
      if (i == 0) {
        cout << 0;
        return 0;
      }
      if (a[i - 1] == b[i - 1] && a[i - 1]) {
        cout << 0;
        return 0;
      }
      if (a[i] == 0) {
        ans += ans;
      }
    } else {
      if (i != 0 && a[i - 1] != b[i - 1] && !a[i - 1]) {
        cout << 0;
        return 0;
      }
      if (a[i] == 1) {
        ans += ans;
      }
    }
  }
  if (second == x) ans -= 2;
  cout << ans;
  return 0;
}

```