#include <bits/stdc++.h>
using namespace std;
bool check(long long n, long long pos) {
  return (n & (((long long)1) << pos)) >> pos;
}
const long long M = 100010;
long long sum, xr, n, len, dp[100][5];
long long call(long long i, long long carry) {
  if (i == n) return carry == 0 && len <= i;
  if (dp[i][carry] != -1ll) return dp[i][carry];
  long long a = 0, b = 0;
  if (check(xr, i) == 0) {
    if (check(sum, i) == 1) {
      if (carry == 1) {
        a = call(i + 1, 0);
        b = call(i + 1, 1);
      }
    } else {
      if (carry == 0) {
        a = call(i + 1, 0);
        b = call(i + 1, 1);
      }
    }
  } else {
    if (check(sum, i) == 1) {
      if (carry == 0) {
        a = call(i + 1, 0);
        b = call(i + 1, 0);
      }
    } else {
      if (carry == 1) {
        a = call(i + 1, 1);
        b = call(i + 1, 1);
      }
    }
  }
  return dp[i][carry] = a + b;
}
int main() {
  long long i, j, k;
  scanf("%I64d%I64d", &sum, &xr);
  memset((dp), (-1ll), sizeof(dp));
  for (i = 60; i >= 0; i--)
    if (check(sum, i)) break;
  n = i + 1;
  for (i = 60; i >= 0; i--)
    if (check(xr, i)) break;
  len = i + 1;
  k = call(0, 0);
  sum == xr ? printf("%I64d\n", max(0ll, k - 2)) : printf("%I64d\n", k);
  return 0;
}

### Prompt

Please create a solution in cpp to 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;
bool check(long long n, long long pos) {
  return (n & (((long long)1) << pos)) >> pos;
}
const long long M = 100010;
long long sum, xr, n, len, dp[100][5];
long long call(long long i, long long carry) {
  if (i == n) return carry == 0 && len <= i;
  if (dp[i][carry] != -1ll) return dp[i][carry];
  long long a = 0, b = 0;
  if (check(xr, i) == 0) {
    if (check(sum, i) == 1) {
      if (carry == 1) {
        a = call(i + 1, 0);
        b = call(i + 1, 1);
      }
    } else {
      if (carry == 0) {
        a = call(i + 1, 0);
        b = call(i + 1, 1);
      }
    }
  } else {
    if (check(sum, i) == 1) {
      if (carry == 0) {
        a = call(i + 1, 0);
        b = call(i + 1, 0);
      }
    } else {
      if (carry == 1) {
        a = call(i + 1, 1);
        b = call(i + 1, 1);
      }
    }
  }
  return dp[i][carry] = a + b;
}
int main() {
  long long i, j, k;
  scanf("%I64d%I64d", &sum, &xr);
  memset((dp), (-1ll), sizeof(dp));
  for (i = 60; i >= 0; i--)
    if (check(sum, i)) break;
  n = i + 1;
  for (i = 60; i >= 0; i--)
    if (check(xr, i)) break;
  len = i + 1;
  k = call(0, 0);
  sum == xr ? printf("%I64d\n", max(0ll, k - 2)) : printf("%I64d\n", k);
  return 0;
}

```