#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int LMAX = 50;
long long s, x;
long long dp[LMAX][2] = {0};
void clear() { memset(dp, 0, sizeof(dp)); }
void solve() {
  dp[0][0] = 1;
  for (int i = 1; i < LMAX; i++) {
    int bx = x >> i - 1 & 1, bs = s >> i - 1 & 1;
    if (bx == 1 && bs == 1) dp[i][1] = 0, dp[i][0] = 2 * dp[i - 1][0];
    if (bx == 1 && bs == 0) dp[i][1] = 2 * dp[i - 1][1], dp[i][0] = 0;
    if (bx == 0 && bs == 0) dp[i][1] = dp[i - 1][0], dp[i][0] = dp[i - 1][0];
    if (bx == 0 && bs == 1) dp[i][1] = dp[i - 1][1], dp[i][0] = dp[i - 1][1];
  }
  long long res = dp[LMAX - 1][0] - (s == x ? 2 : 0);
  printf("%lld\n", res);
}
int main() {
  while (scanf("%lld%lld", &s, &x) == 2) {
    clear();
    solve();
    return 0;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int LMAX = 50;
long long s, x;
long long dp[LMAX][2] = {0};
void clear() { memset(dp, 0, sizeof(dp)); }
void solve() {
  dp[0][0] = 1;
  for (int i = 1; i < LMAX; i++) {
    int bx = x >> i - 1 & 1, bs = s >> i - 1 & 1;
    if (bx == 1 && bs == 1) dp[i][1] = 0, dp[i][0] = 2 * dp[i - 1][0];
    if (bx == 1 && bs == 0) dp[i][1] = 2 * dp[i - 1][1], dp[i][0] = 0;
    if (bx == 0 && bs == 0) dp[i][1] = dp[i - 1][0], dp[i][0] = dp[i - 1][0];
    if (bx == 0 && bs == 1) dp[i][1] = dp[i - 1][1], dp[i][0] = dp[i - 1][1];
  }
  long long res = dp[LMAX - 1][0] - (s == x ? 2 : 0);
  printf("%lld\n", res);
}
int main() {
  while (scanf("%lld%lld", &s, &x) == 2) {
    clear();
    solve();
    return 0;
  }
  return 0;
}

```