#include <bits/stdc++.h>
using namespace std;
const int N = 400005;
long long dp[100][5];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  long long ans = 0;
  if (s == x) {
    ans = -2LL;
  }
  vector<int> vs, vx;
  while (s) {
    vs.push_back(s % 2LL);
    s /= 2LL;
  }
  while (x) {
    vx.push_back(x % 2LL);
    x /= 2LL;
  }
  if (vx.size() < vs.size()) {
    while (vx.size() < vs.size()) {
      vx.push_back(0);
    }
  } else {
    while (vs.size() < vx.size()) {
      vs.push_back(0);
    }
  }
  int sz = vx.size();
  dp[0][0] = 1;
  for (int i = 0; i < sz; i++) {
    for (int c = 0; c <= 1; c++) {
      for (int a = 0; a <= 1; a++) {
        for (int b = 0; b <= 1; b++) {
          int sum = a + b + c;
          int dd = sum % 2;
          int cc = sum / 2;
          if ((a ^ b == vx[i]) && (dd == vs[i])) {
            dp[i + 1][cc] += dp[i][c];
          }
        }
      }
    }
  }
  ans += dp[sz][0];
  if (ans < 0LL) {
    ans = 0LL;
  }
  cout << ans << "\n";
}

### Prompt

Your challenge is to write 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 N = 400005;
long long dp[100][5];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  long long ans = 0;
  if (s == x) {
    ans = -2LL;
  }
  vector<int> vs, vx;
  while (s) {
    vs.push_back(s % 2LL);
    s /= 2LL;
  }
  while (x) {
    vx.push_back(x % 2LL);
    x /= 2LL;
  }
  if (vx.size() < vs.size()) {
    while (vx.size() < vs.size()) {
      vx.push_back(0);
    }
  } else {
    while (vs.size() < vx.size()) {
      vs.push_back(0);
    }
  }
  int sz = vx.size();
  dp[0][0] = 1;
  for (int i = 0; i < sz; i++) {
    for (int c = 0; c <= 1; c++) {
      for (int a = 0; a <= 1; a++) {
        for (int b = 0; b <= 1; b++) {
          int sum = a + b + c;
          int dd = sum % 2;
          int cc = sum / 2;
          if ((a ^ b == vx[i]) && (dd == vs[i])) {
            dp[i + 1][cc] += dp[i][c];
          }
        }
      }
    }
  }
  ans += dp[sz][0];
  if (ans < 0LL) {
    ans = 0LL;
  }
  cout << ans << "\n";
}

```