#include <bits/stdc++.h>
using namespace std;
long long s, x;
bool bitX[40], bitS[40];
long long dp[41][2];
int maxBit;
long long solve(int bit, bool carry) {
  if (bit == maxBit) return !carry;
  if (dp[bit][carry] != -1) return dp[bit][carry];
  long long ans;
  if (bitX[bit]) {
    if (carry) {
      if (bitS[bit])
        ans = 0;
      else
        ans = 2LL * solve(bit + 1, 1);
    } else {
      if (!bitS[bit])
        ans = 0;
      else
        ans = 2LL * solve(bit + 1, 0);
    }
  } else {
    if (carry) {
      if (!bitS[bit])
        ans = 0;
      else
        ans = solve(bit + 1, 0) + solve(bit + 1, 1);
    } else {
      if (bitS[bit])
        ans = 0;
      else
        ans = solve(bit + 1, 0) + solve(bit + 1, 1);
    }
  }
  return dp[bit][carry] = ans;
}
int main() {
  cin >> s >> x;
  long long temp = s, temp2 = x;
  int neg = x == s ? -2 : 0;
  while (temp || temp2) {
    maxBit++;
    temp /= 2;
    temp2 /= 2;
  }
  for (long long i = 0; i < 40; i++) {
    bitX[i] = x % 2;
    x /= 2;
    bitS[i] = s % 2;
    s /= 2;
  }
  memset(dp, -1, sizeof dp);
  long long ans = solve(0, 0);
  cout << ans + neg;
  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;
bool bitX[40], bitS[40];
long long dp[41][2];
int maxBit;
long long solve(int bit, bool carry) {
  if (bit == maxBit) return !carry;
  if (dp[bit][carry] != -1) return dp[bit][carry];
  long long ans;
  if (bitX[bit]) {
    if (carry) {
      if (bitS[bit])
        ans = 0;
      else
        ans = 2LL * solve(bit + 1, 1);
    } else {
      if (!bitS[bit])
        ans = 0;
      else
        ans = 2LL * solve(bit + 1, 0);
    }
  } else {
    if (carry) {
      if (!bitS[bit])
        ans = 0;
      else
        ans = solve(bit + 1, 0) + solve(bit + 1, 1);
    } else {
      if (bitS[bit])
        ans = 0;
      else
        ans = solve(bit + 1, 0) + solve(bit + 1, 1);
    }
  }
  return dp[bit][carry] = ans;
}
int main() {
  cin >> s >> x;
  long long temp = s, temp2 = x;
  int neg = x == s ? -2 : 0;
  while (temp || temp2) {
    maxBit++;
    temp /= 2;
    temp2 /= 2;
  }
  for (long long i = 0; i < 40; i++) {
    bitX[i] = x % 2;
    x /= 2;
    bitS[i] = s % 2;
    s /= 2;
  }
  memset(dp, -1, sizeof dp);
  long long ans = solve(0, 0);
  cout << ans + neg;
  return 0;
}

```