#include <bits/stdc++.h>
using namespace std;
const int iinf = 1e9 + 7;
const long long int inf = 1e9 + 7;
const long long int mod = 998244353;
const long long int maxn = 2e5 + 5;
long long int sqr(long long int x) { return x * x; }
long long int modd(long long int a) {
  if (a > 0) return a % inf;
  while (a < 0) {
    a += inf;
  }
  return a;
}
long long int modexpo(long long int a, long long int b) {
  a %= inf;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % inf;
    a = a * a % inf;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long int x, y;
  cin >> x >> y;
  long long int z = x - y;
  cout << (z < 0 || z & 1 || (z / 2) & y
               ? 0
               : (1ll << __builtin_popcountll(y)) - 2 * (x == y));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Generate a cpp solution 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;
const int iinf = 1e9 + 7;
const long long int inf = 1e9 + 7;
const long long int mod = 998244353;
const long long int maxn = 2e5 + 5;
long long int sqr(long long int x) { return x * x; }
long long int modd(long long int a) {
  if (a > 0) return a % inf;
  while (a < 0) {
    a += inf;
  }
  return a;
}
long long int modexpo(long long int a, long long int b) {
  a %= inf;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % inf;
    a = a * a % inf;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long int x, y;
  cin >> x >> y;
  long long int z = x - y;
  cout << (z < 0 || z & 1 || (z / 2) & y
               ? 0
               : (1ll << __builtin_popcountll(y)) - 2 * (x == y));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```