#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long dp[60][2];
long long rek(int cbit, int carry) {
  if (cbit == 60) return carry == 0;
  if (dp[cbit][carry] == -1) {
    long long &rez = dp[cbit][carry];
    rez = 0;
    int need = -carry;
    if (s & (1LL << cbit)) {
      need++;
    }
    int iks = (x & (1LL << cbit)) > 0;
    if (need == -1 && iks == 1) {
      rez = 2 * rek(cbit + 1, 1);
    }
    if (need == 0 && iks == 0) {
      rez = rek(cbit + 1, 0) + rek(cbit + 1, 1);
    } else if (need == 1 && iks == 1) {
      rez = 2 * rek(cbit + 1, 0);
    }
  }
  return dp[cbit][carry];
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> s >> x;
  cout << rek(0, 0) - 2 * (s == x);
  return 0;
}

### 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;
long long s, x;
long long dp[60][2];
long long rek(int cbit, int carry) {
  if (cbit == 60) return carry == 0;
  if (dp[cbit][carry] == -1) {
    long long &rez = dp[cbit][carry];
    rez = 0;
    int need = -carry;
    if (s & (1LL << cbit)) {
      need++;
    }
    int iks = (x & (1LL << cbit)) > 0;
    if (need == -1 && iks == 1) {
      rez = 2 * rek(cbit + 1, 1);
    }
    if (need == 0 && iks == 0) {
      rez = rek(cbit + 1, 0) + rek(cbit + 1, 1);
    } else if (need == 1 && iks == 1) {
      rez = 2 * rek(cbit + 1, 0);
    }
  }
  return dp[cbit][carry];
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> s >> x;
  cout << rek(0, 0) - 2 * (s == x);
  return 0;
}

```