#include <bits/stdc++.h>
using namespace std;
const int N = 61;
long long dp[N][2];
long long a, b;
int getBit(long long x, int j) {
  if ((x & (1ll << j)))
    return 1;
  else
    return 0;
}
int main() {
  memset(dp, 0, sizeof(dp));
  scanf("%I64d%I64d", &a, &b);
  dp[60][0] = 1;
  for (int i = 59; i >= 0; --i) {
    int c1 = getBit(a, i);
    int c2 = getBit(b, i);
    if (c1 == 1 && c2 == 1) {
      dp[i][0] = 2 * dp[i + 1][0];
      dp[i][1] = 0;
    }
    if (c1 == 0 && c2 == 1) {
      dp[i][0] = 0;
      dp[i][1] = 2 * dp[i + 1][1];
    }
    if (c1 == 1 && c2 == 0) {
      dp[i][0] = 0;
      dp[i][1] = dp[i + 1][0] + dp[i + 1][1];
    }
    if (c1 == 0 && c2 == 0) {
      dp[i][0] = dp[i + 1][0] + dp[i + 1][1];
      dp[i][1] = 0;
    }
  }
  if (a == b)
    cout << dp[0][0] - 2;
  else
    cout << dp[0][0];
  return 0;
}

### 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;
const int N = 61;
long long dp[N][2];
long long a, b;
int getBit(long long x, int j) {
  if ((x & (1ll << j)))
    return 1;
  else
    return 0;
}
int main() {
  memset(dp, 0, sizeof(dp));
  scanf("%I64d%I64d", &a, &b);
  dp[60][0] = 1;
  for (int i = 59; i >= 0; --i) {
    int c1 = getBit(a, i);
    int c2 = getBit(b, i);
    if (c1 == 1 && c2 == 1) {
      dp[i][0] = 2 * dp[i + 1][0];
      dp[i][1] = 0;
    }
    if (c1 == 0 && c2 == 1) {
      dp[i][0] = 0;
      dp[i][1] = 2 * dp[i + 1][1];
    }
    if (c1 == 1 && c2 == 0) {
      dp[i][0] = 0;
      dp[i][1] = dp[i + 1][0] + dp[i + 1][1];
    }
    if (c1 == 0 && c2 == 0) {
      dp[i][0] = dp[i + 1][0] + dp[i + 1][1];
      dp[i][1] = 0;
    }
  }
  if (a == b)
    cout << dp[0][0] - 2;
  else
    cout << dp[0][0];
  return 0;
}

```