#include <bits/stdc++.h>
using namespace std;
long long solve(long long s, long long x) {
  if ((s - x) % 2 || x > s) {
    return 0;
  }
  vector<int> sb(64, 0), xb(64, 0), sxb(64, 0);
  long long t = s, k = 0;
  while (t > 0) {
    sb[k] = t % 2;
    t /= 2;
    k++;
  }
  t = x;
  k = 0;
  while (t > 0) {
    xb[k] = t % 2;
    t /= 2;
    k++;
  }
  t = (s - x) / 2;
  k = 0;
  while (t > 0) {
    sxb[k] = t % 2;
    t /= 2;
    k++;
  }
  long long wyn = 0;
  for (int i = 0; i < 64; i++) {
    wyn += xb[i];
  }
  wyn = pow(2, wyn);
  if (s == x) wyn -= 2;
  bool spr = 1;
  for (int i = 0; i < 64; i++)
    if (sxb[i] + xb[i] == 2) spr = 0;
  if (spr) return wyn;
  ;
  return 0;
}
long long brut(long long s, long long x) {
  long long wyn = 0;
  for (int i = 1; i < s; i++) {
    if ((i ^ (s - i)) == x) wyn++;
  }
  return wyn;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x);
  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;
long long solve(long long s, long long x) {
  if ((s - x) % 2 || x > s) {
    return 0;
  }
  vector<int> sb(64, 0), xb(64, 0), sxb(64, 0);
  long long t = s, k = 0;
  while (t > 0) {
    sb[k] = t % 2;
    t /= 2;
    k++;
  }
  t = x;
  k = 0;
  while (t > 0) {
    xb[k] = t % 2;
    t /= 2;
    k++;
  }
  t = (s - x) / 2;
  k = 0;
  while (t > 0) {
    sxb[k] = t % 2;
    t /= 2;
    k++;
  }
  long long wyn = 0;
  for (int i = 0; i < 64; i++) {
    wyn += xb[i];
  }
  wyn = pow(2, wyn);
  if (s == x) wyn -= 2;
  bool spr = 1;
  for (int i = 0; i < 64; i++)
    if (sxb[i] + xb[i] == 2) spr = 0;
  if (spr) return wyn;
  ;
  return 0;
}
long long brut(long long s, long long x) {
  long long wyn = 0;
  for (int i = 1; i < s; i++) {
    if ((i ^ (s - i)) == x) wyn++;
  }
  return wyn;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x);
  return 0;
}

```