#include <bits/stdc++.h>
using namespace std;
const int MAXN = 400000;
const int MAXM = 200000;
const int INF = 1000000010;
const long long int LLINF = 1000000000000000010LL;
const long long int MOD = 1000000007;
const long long int P = 31;
const double EPS = 1e-6;
long long int S, X;
long long int dp[64][2] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  scanf("%I64d %I64d", &S, &X);
  if ((S & 1) == 0) {
    if ((X & 1) == 0) {
      dp[0][0] = dp[0][1] = 1;
    }
  } else {
    if ((X & 1) == 1) {
      dp[0][0] = 2;
    }
  }
  int bi;
  for (int i = 1; i < 64; i++) {
    for (int ai = 0; ai < 2; ai++) {
      bi = ((X >> i) ^ ai) & 1;
      if (ai + bi == ((S >> i) & 1) + 2) {
        dp[i][1] += dp[i - 1][0];
      }
      if (ai + bi + 1 == ((S >> i) & 1) + 2) {
        dp[i][1] += dp[i - 1][1];
      }
      if (ai + bi + 1 == ((S >> i) & 1)) {
        dp[i][0] += dp[i - 1][1];
      }
      if (ai + bi == ((S >> i) & 1)) {
        dp[i][0] += dp[i - 1][0];
      }
    }
  }
  if (0 + X == S) {
    dp[63][0] -= 2;
  }
  printf("%I64d", dp[63][0]);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 int MAXN = 400000;
const int MAXM = 200000;
const int INF = 1000000010;
const long long int LLINF = 1000000000000000010LL;
const long long int MOD = 1000000007;
const long long int P = 31;
const double EPS = 1e-6;
long long int S, X;
long long int dp[64][2] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  scanf("%I64d %I64d", &S, &X);
  if ((S & 1) == 0) {
    if ((X & 1) == 0) {
      dp[0][0] = dp[0][1] = 1;
    }
  } else {
    if ((X & 1) == 1) {
      dp[0][0] = 2;
    }
  }
  int bi;
  for (int i = 1; i < 64; i++) {
    for (int ai = 0; ai < 2; ai++) {
      bi = ((X >> i) ^ ai) & 1;
      if (ai + bi == ((S >> i) & 1) + 2) {
        dp[i][1] += dp[i - 1][0];
      }
      if (ai + bi + 1 == ((S >> i) & 1) + 2) {
        dp[i][1] += dp[i - 1][1];
      }
      if (ai + bi + 1 == ((S >> i) & 1)) {
        dp[i][0] += dp[i - 1][1];
      }
      if (ai + bi == ((S >> i) & 1)) {
        dp[i][0] += dp[i - 1][0];
      }
    }
  }
  if (0 + X == S) {
    dp[63][0] -= 2;
  }
  printf("%I64d", dp[63][0]);
  return 0;
}

```