#include <bits/stdc++.h>
using namespace std;
const long long N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
long long s, x, dp[111][5][5][5];
long long calc(long long pos, long long ost, long long was, long long was2) {
  if (pos == 50ll) {
    return (!ost && was && was2);
  }
  long long &res = dp[pos][ost][was][was2];
  if (~res) return res;
  res = 0;
  for (long long i = 0; i <= 1; ++i) {
    for (long long j = 0; j <= 1; ++j) {
      if (((i + j + ost) % 2ll) == ((bool)((1ll << pos) & s)) &&
          (i ^ j) == ((bool)((1ll << pos) & x))) {
        res += calc(pos + 1, (i + j + ost) / 2ll, was | i, was2 | j);
      }
    }
  }
  return res;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> s >> x;
  memset(dp, -1, sizeof dp);
  cout << calc(0ll, 0ll, 0ll, 0ll);
  return 0;
}

### Prompt

Please formulate 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 long long N = (long long)(5e5) + 322;
const long long INF = (long long)(1e9);
const long long mod = (long long)(1e9) + 7;
const double eps = 1e-9;
long long s, x, dp[111][5][5][5];
long long calc(long long pos, long long ost, long long was, long long was2) {
  if (pos == 50ll) {
    return (!ost && was && was2);
  }
  long long &res = dp[pos][ost][was][was2];
  if (~res) return res;
  res = 0;
  for (long long i = 0; i <= 1; ++i) {
    for (long long j = 0; j <= 1; ++j) {
      if (((i + j + ost) % 2ll) == ((bool)((1ll << pos) & s)) &&
          (i ^ j) == ((bool)((1ll << pos) & x))) {
        res += calc(pos + 1, (i + j + ost) / 2ll, was | i, was2 | j);
      }
    }
  }
  return res;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> s >> x;
  memset(dp, -1, sizeof dp);
  cout << calc(0ll, 0ll, 0ll, 0ll);
  return 0;
}

```