#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100;
long long s, x;
long long a, b;
long long memes[MAXN + 3][MAXN + 3];
long long rec(int pos, int carry) {
  if (memes[pos][carry] != -1) {
    return memes[pos][carry];
  }
  int xd = (x >> pos) & 1;
  int sd = (s >> pos) & 1;
  if ((x >> pos) == 0 && (s >> pos) == 0 && carry == 0) {
    return 1;
  }
  long long ret = 0;
  for (int i = 0; i <= 1; i++) {
    for (int j = 0; j <= 1; j++) {
      if (i ^ j == xd && (i + j + carry) % 2 == sd) {
        ret += rec(pos + 1, i + j + carry >= 2 ? 1 : 0);
      }
    }
  }
  memes[pos][carry] = ret;
  return ret;
}
int main(int argc, char* argv[]) {
  scanf("%I64d %I64d", &s, &x);
  fill(&memes[0][0], &memes[0][0] + sizeof(memes) / sizeof(long long), -1);
  long long ans = rec(0, 0);
  if (s == x) {
    ans -= 2;
  }
  printf("%I64d\n", ans);
  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 MAXN = 100;
long long s, x;
long long a, b;
long long memes[MAXN + 3][MAXN + 3];
long long rec(int pos, int carry) {
  if (memes[pos][carry] != -1) {
    return memes[pos][carry];
  }
  int xd = (x >> pos) & 1;
  int sd = (s >> pos) & 1;
  if ((x >> pos) == 0 && (s >> pos) == 0 && carry == 0) {
    return 1;
  }
  long long ret = 0;
  for (int i = 0; i <= 1; i++) {
    for (int j = 0; j <= 1; j++) {
      if (i ^ j == xd && (i + j + carry) % 2 == sd) {
        ret += rec(pos + 1, i + j + carry >= 2 ? 1 : 0);
      }
    }
  }
  memes[pos][carry] = ret;
  return ret;
}
int main(int argc, char* argv[]) {
  scanf("%I64d %I64d", &s, &x);
  fill(&memes[0][0], &memes[0][0] + sizeof(memes) / sizeof(long long), -1);
  long long ans = rec(0, 0);
  if (s == x) {
    ans -= 2;
  }
  printf("%I64d\n", ans);
  return 0;
}

```