#include <bits/stdc++.h>
using namespace std;
constexpr static int N = 40;
long long cnt[N][2];
long long solve() {
  long long s, x;
  cin >> s >> x;
  cnt[0][0] = 0;
  cnt[0][1] = 0;
  if ((s & 1) == (x & 1)) {
    if (s & 1)
      cnt[0][0] = 2;
    else
      cnt[0][0] = cnt[0][1] = 1;
  } else {
    return 0;
  }
  for (int i = 1; i < N; i++) {
    cnt[i][0] = 0;
    cnt[i][1] = 0;
    bool sbit = (s & (1LL << i)) != 0;
    bool xbit = (x & (1LL << i)) != 0;
    if (sbit && xbit) {
      cnt[i][0] += 2 * cnt[i - 1][0];
    }
    if (sbit && !xbit) {
      cnt[i][0] += cnt[i - 1][1];
      cnt[i][1] += cnt[i - 1][1];
    }
    if (!sbit && xbit) {
      cnt[i][1] += 2 * cnt[i - 1][1];
    }
    if (!sbit && !xbit) {
      cnt[i][0] += cnt[i - 1][0];
      cnt[i][1] += cnt[i - 1][0];
    }
  }
  long long r = cnt[N - 1][0];
  if (x == s) r -= 2;
  return r;
}
int main() { cout << solve() << endl; }

### 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;
constexpr static int N = 40;
long long cnt[N][2];
long long solve() {
  long long s, x;
  cin >> s >> x;
  cnt[0][0] = 0;
  cnt[0][1] = 0;
  if ((s & 1) == (x & 1)) {
    if (s & 1)
      cnt[0][0] = 2;
    else
      cnt[0][0] = cnt[0][1] = 1;
  } else {
    return 0;
  }
  for (int i = 1; i < N; i++) {
    cnt[i][0] = 0;
    cnt[i][1] = 0;
    bool sbit = (s & (1LL << i)) != 0;
    bool xbit = (x & (1LL << i)) != 0;
    if (sbit && xbit) {
      cnt[i][0] += 2 * cnt[i - 1][0];
    }
    if (sbit && !xbit) {
      cnt[i][0] += cnt[i - 1][1];
      cnt[i][1] += cnt[i - 1][1];
    }
    if (!sbit && xbit) {
      cnt[i][1] += 2 * cnt[i - 1][1];
    }
    if (!sbit && !xbit) {
      cnt[i][0] += cnt[i - 1][0];
      cnt[i][1] += cnt[i - 1][0];
    }
  }
  long long r = cnt[N - 1][0];
  if (x == s) r -= 2;
  return r;
}
int main() { cout << solve() << endl; }

```