#include <bits/stdc++.h>
using namespace std;
const int maxn = 20;
inline void read(long long &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void judge() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
map<pair<pair<long long, long long>, bool>, long long> f;
inline long long dp(long long s, long long x, bool jin) {
  if (f.count(make_pair(make_pair(s, x), jin)))
    return f[make_pair(make_pair(s, x), jin)];
  long long &res = f[make_pair(make_pair(s, x), jin)];
  if (s == 0 && x == 0) {
    if (jin)
      return res = 0;
    else
      return res = 1;
  }
  int v1 = s & 1, v2 = x & 1;
  for (int i = (0); i <= (1); i++)
    for (int j = (0); j <= (1); j++) {
      if ((((i + j + jin) & 1) == v1) && ((i xor j) == v2))
        res += dp(s >> 1, x >> 1, (i + j + jin) >> 1);
    }
  return res;
}
long long s, x;
int main() {
  read(s);
  read(x);
  cout << dp(s, x, 0) - (s == x) * 2;
  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;
const int maxn = 20;
inline void read(long long &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void judge() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
map<pair<pair<long long, long long>, bool>, long long> f;
inline long long dp(long long s, long long x, bool jin) {
  if (f.count(make_pair(make_pair(s, x), jin)))
    return f[make_pair(make_pair(s, x), jin)];
  long long &res = f[make_pair(make_pair(s, x), jin)];
  if (s == 0 && x == 0) {
    if (jin)
      return res = 0;
    else
      return res = 1;
  }
  int v1 = s & 1, v2 = x & 1;
  for (int i = (0); i <= (1); i++)
    for (int j = (0); j <= (1); j++) {
      if ((((i + j + jin) & 1) == v1) && ((i xor j) == v2))
        res += dp(s >> 1, x >> 1, (i + j + jin) >> 1);
    }
  return res;
}
long long s, x;
int main() {
  read(s);
  read(x);
  cout << dp(s, x, 0) - (s == x) * 2;
  return 0;
}

```