#include <bits/stdc++.h>
using namespace std;
using lli = long long;
int s[60], v[60], n;
lli dp[60][2][2][2];
lli DP(int idx, int t, int af, int bf) {
  lli &ret = dp[idx][t][af][bf];
  if (ret != -1) return ret;
  if (idx == n) return ret = ((t == 0) && af && bf);
  ret = 0;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      if ((i ^ j) == v[idx] && (i + j + t) % 2 == s[idx]) {
        int ns = (t + i + j) >= 2;
        ret += DP(idx + 1, ns, af | i, bf | j);
      }
  return ret;
}
int main() {
  lli a, b;
  scanf("%lld %lld", &a, &b);
  while (a || b) {
    s[n] = a % 2;
    v[n] = b % 2;
    n++;
    a >>= 1;
    b >>= 1;
  }
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", DP(0, 0, 0, 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;
using lli = long long;
int s[60], v[60], n;
lli dp[60][2][2][2];
lli DP(int idx, int t, int af, int bf) {
  lli &ret = dp[idx][t][af][bf];
  if (ret != -1) return ret;
  if (idx == n) return ret = ((t == 0) && af && bf);
  ret = 0;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      if ((i ^ j) == v[idx] && (i + j + t) % 2 == s[idx]) {
        int ns = (t + i + j) >= 2;
        ret += DP(idx + 1, ns, af | i, bf | j);
      }
  return ret;
}
int main() {
  lli a, b;
  scanf("%lld %lld", &a, &b);
  while (a || b) {
    s[n] = a % 2;
    v[n] = b % 2;
    n++;
    a >>= 1;
    b >>= 1;
  }
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", DP(0, 0, 0, 0));
  return 0;
}

```