#include <bits/stdc++.h>
using namespace std;
long long s, x;
int bits, bs[45], bx[45];
long long dp[45][2];
long long go(int bit, int carry) {
  if (dp[bit][carry] != -1) return dp[bit][carry];
  if (bit == bits) return dp[bit][carry] = carry == 0;
  long long ret = 0;
  if (carry == 0) {
    if (bx[bit] == bs[bit]) {
      ret += go(bit + 1, carry);
      if (bx[bit] == 0)
        ret += go(bit + 1, carry ^ 1);
      else
        ret += go(bit + 1, carry);
    }
  } else {
    if (bx[bit] != bs[bit]) {
      if (bx[bit] == 0)
        ret += go(bit + 1, carry ^ 1);
      else
        ret += go(bit + 1, carry);
      ret += go(bit + 1, carry);
    }
  }
  return dp[bit][carry] = ret;
}
int main() {
  cin >> s >> x;
  bits = max(64 - __builtin_clzll(s), 64 - __builtin_clzll(x));
  for (int i = 0; i < 45; i++) bs[i] = (s >> i) & 1;
  for (int i = 0; i < 45; i++) bx[i] = (x >> i) & 1;
  memset(dp, -1, sizeof(dp));
  long long ans = go(0, 0);
  if (s == x) ans -= 2;
  cout << ans << 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;
long long s, x;
int bits, bs[45], bx[45];
long long dp[45][2];
long long go(int bit, int carry) {
  if (dp[bit][carry] != -1) return dp[bit][carry];
  if (bit == bits) return dp[bit][carry] = carry == 0;
  long long ret = 0;
  if (carry == 0) {
    if (bx[bit] == bs[bit]) {
      ret += go(bit + 1, carry);
      if (bx[bit] == 0)
        ret += go(bit + 1, carry ^ 1);
      else
        ret += go(bit + 1, carry);
    }
  } else {
    if (bx[bit] != bs[bit]) {
      if (bx[bit] == 0)
        ret += go(bit + 1, carry ^ 1);
      else
        ret += go(bit + 1, carry);
      ret += go(bit + 1, carry);
    }
  }
  return dp[bit][carry] = ret;
}
int main() {
  cin >> s >> x;
  bits = max(64 - __builtin_clzll(s), 64 - __builtin_clzll(x));
  for (int i = 0; i < 45; i++) bs[i] = (s >> i) & 1;
  for (int i = 0; i < 45; i++) bx[i] = (x >> i) & 1;
  memset(dp, -1, sizeof(dp));
  long long ans = go(0, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
}

```