#include <bits/stdc++.h>
using namespace std;
long long int s, x, dp[100][10][10][10], ar[100][100];
long long int solve(int c, int d, int a, int b) {
  if (c > 50) {
    if (a && b && !d) return 1;
    return 0;
  }
  long long int &ret = dp[c][d][a][b];
  if (ret != -1) return ret;
  ret = 0;
  if (ar[1][c] == 0 && d % 2 == ar[0][c])
    ret += solve(c + 1, d / 2, a | 0, b | 0);
  if (ar[1][c] == 1 && (1 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 1) / 2, a | 1, b | 0);
  if (ar[1][c] == 1 && (1 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 1) / 2, a | 0, b | 1);
  if (ar[1][c] == 0 && (2 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 2) / 2, a | 1, b | 1);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> x;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i <= 51; i++) {
    if (s & (1ll << i)) ar[0][i] = 1;
    if (x & (1ll << i)) ar[1][i] = 1;
  }
  cout << solve(0, 0, 0, 0);
}

### Prompt

Please provide a CPP coded solution 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;
long long int s, x, dp[100][10][10][10], ar[100][100];
long long int solve(int c, int d, int a, int b) {
  if (c > 50) {
    if (a && b && !d) return 1;
    return 0;
  }
  long long int &ret = dp[c][d][a][b];
  if (ret != -1) return ret;
  ret = 0;
  if (ar[1][c] == 0 && d % 2 == ar[0][c])
    ret += solve(c + 1, d / 2, a | 0, b | 0);
  if (ar[1][c] == 1 && (1 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 1) / 2, a | 1, b | 0);
  if (ar[1][c] == 1 && (1 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 1) / 2, a | 0, b | 1);
  if (ar[1][c] == 0 && (2 + d) % 2 == ar[0][c])
    ret += solve(c + 1, (d + 2) / 2, a | 1, b | 1);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> x;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i <= 51; i++) {
    if (s & (1ll << i)) ar[0][i] = 1;
    if (x & (1ll << i)) ar[1][i] = 1;
  }
  cout << solve(0, 0, 0, 0);
}

```