#include <bits/stdc++.h>
using namespace std;
long long int bpow(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int res = bpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
long long int bpowm(long long int x, long long int y, long long int m) {
  long long int res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
long long int solve() {
  long long int s, x;
  cin >> s >> x;
  long long int k = (s - x) / 2;
  if ((s - x) % 2 != 0)
    cout << 0 << "\n";
  else {
    long long int count = 0;
    for (long long int i = 0; i < 41; i++) {
      if ((x & ((long long)1 << i))) {
        if (k & ((long long)1 << i)) {
          cout << 0;
          return 0;
        }
        count++;
      }
    }
    if (x != s)
      cout << bpow(2, count);
    else
      cout << bpow(2, count) - 2;
  }
  return 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 int bpow(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int res = bpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
long long int bpowm(long long int x, long long int y, long long int m) {
  long long int res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
long long int solve() {
  long long int s, x;
  cin >> s >> x;
  long long int k = (s - x) / 2;
  if ((s - x) % 2 != 0)
    cout << 0 << "\n";
  else {
    long long int count = 0;
    for (long long int i = 0; i < 41; i++) {
      if ((x & ((long long)1 << i))) {
        if (k & ((long long)1 << i)) {
          cout << 0;
          return 0;
        }
        count++;
      }
    }
    if (x != s)
      cout << bpow(2, count);
    else
      cout << bpow(2, count) - 2;
  }
  return 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```