#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  vector<pair<long long, long long> > dp(42);
  for (long long i = 0; i < 42; ++i) {
    int s_ = int(bool(s & (1ll << i)));
    int x_ = int(bool(x & (1ll << i)));
    long long p1 = (i ? dp[i - 1].first : 0);
    long long p0 = (i ? dp[i - 1].second : 1);
    if (s_ == 1 && x_ == 1) dp[i] = make_pair(0, 2 * p0);
    if (s_ == 1 && x_ == 0) dp[i] = make_pair(p1, p1);
    if (s_ == 0 && x_ == 1) dp[i] = make_pair(2 * p1, 0);
    if (s_ == 0 && x_ == 0) dp[i] = make_pair(p0, p0);
  }
  if (x == s) dp.back().second -= 2;
  cout << dp.back().second << endl;
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  vector<pair<long long, long long> > dp(42);
  for (long long i = 0; i < 42; ++i) {
    int s_ = int(bool(s & (1ll << i)));
    int x_ = int(bool(x & (1ll << i)));
    long long p1 = (i ? dp[i - 1].first : 0);
    long long p0 = (i ? dp[i - 1].second : 1);
    if (s_ == 1 && x_ == 1) dp[i] = make_pair(0, 2 * p0);
    if (s_ == 1 && x_ == 0) dp[i] = make_pair(p1, p1);
    if (s_ == 0 && x_ == 1) dp[i] = make_pair(2 * p1, 0);
    if (s_ == 0 && x_ == 0) dp[i] = make_pair(p0, p0);
  }
  if (x == s) dp.back().second -= 2;
  cout << dp.back().second << endl;
}

```