#include <bits/stdc++.h>
using namespace std;
const double kEps = 10e-8;
const int kMax = 1000;
const int kInf = 1 << 30;
long long s, x;
int msb;
long long dp[64][2][2][2];
long long f(int p, int c, int a, int b) {
  if (p > msb) {
    return (c == 0 && a == 1 && b == 1) ? 1 : 0;
  }
  long long& ans = dp[p][c][a][b];
  if (ans != -1) {
    return ans;
  }
  ans = 0;
  int sb = (s & (1ll << p)) != 0;
  int xb = (x & (1ll << p)) != 0;
  if (xb == 0) {
    int v = c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, a, b);
    }
    v = 2 + c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, 1, 1);
    }
  } else {
    int v = 1 + c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, 1, b);
      ans += f(p + 1, v / 2, a, 1);
    }
  }
  return ans;
}
int main() {
  cin >> s >> x;
  msb = 0;
  for (int i = 0; i < 63; i++) {
    if (s & (1ll << i)) {
      msb = i;
    }
    if (x & (1ll << i)) {
      msb = i;
    }
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0, 0, 0);
  return EXIT_SUCCESS;
}

### Prompt

Please create a solution in cpp 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 double kEps = 10e-8;
const int kMax = 1000;
const int kInf = 1 << 30;
long long s, x;
int msb;
long long dp[64][2][2][2];
long long f(int p, int c, int a, int b) {
  if (p > msb) {
    return (c == 0 && a == 1 && b == 1) ? 1 : 0;
  }
  long long& ans = dp[p][c][a][b];
  if (ans != -1) {
    return ans;
  }
  ans = 0;
  int sb = (s & (1ll << p)) != 0;
  int xb = (x & (1ll << p)) != 0;
  if (xb == 0) {
    int v = c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, a, b);
    }
    v = 2 + c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, 1, 1);
    }
  } else {
    int v = 1 + c;
    if (v % 2 == sb) {
      ans += f(p + 1, v / 2, 1, b);
      ans += f(p + 1, v / 2, a, 1);
    }
  }
  return ans;
}
int main() {
  cin >> s >> x;
  msb = 0;
  for (int i = 0; i < 63; i++) {
    if (s & (1ll << i)) {
      msb = i;
    }
    if (x & (1ll << i)) {
      msb = i;
    }
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0, 0, 0);
  return EXIT_SUCCESS;
}

```