#include <bits/stdc++.h>
char _;
using namespace std;
const int MAXN = 51, MOD = 1e9 + 7;
long long S, X;
int s[MAXN], x[MAXN], len;
long long dp[MAXN][2][2][2];
long long solve(int i, bool carry, bool f, bool se) {
  if (i == len + 1 && !carry && f && se) return 1;
  if (i == len + 1) return 0;
  long long &ret = dp[i][carry][f][se];
  if (ret != -1) return ret;
  ret = 0;
  if (s[i] && x[i])
    if (!carry) ret += solve(i + 1, 0, 1, se) + solve(i + 1, 0, f, 1);
  if (s[i] && !x[i])
    if (carry) ret += solve(i + 1, 0, f, se) + solve(i + 1, 1, 1, 1);
  if (!s[i] && x[i])
    if (carry) ret += solve(i + 1, 1, 1, se) + solve(i + 1, 1, f, 1);
  if (!s[i] && !x[i])
    if (!carry) ret += solve(i + 1, 1, 1, 1) + solve(i + 1, 0, f, se);
  return ret;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> S >> X;
  if (S < X) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 50; i++) {
    if ((long long)(1ULL << i) > S) break;
    s[i] = (S & (long long)(1ULL << i)) ? 1 : 0;
    x[i] = (X & (long long)(1ULL << i)) ? 1 : 0;
    if (s[i]) len = i;
  }
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0, 0, 0) << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
char _;
using namespace std;
const int MAXN = 51, MOD = 1e9 + 7;
long long S, X;
int s[MAXN], x[MAXN], len;
long long dp[MAXN][2][2][2];
long long solve(int i, bool carry, bool f, bool se) {
  if (i == len + 1 && !carry && f && se) return 1;
  if (i == len + 1) return 0;
  long long &ret = dp[i][carry][f][se];
  if (ret != -1) return ret;
  ret = 0;
  if (s[i] && x[i])
    if (!carry) ret += solve(i + 1, 0, 1, se) + solve(i + 1, 0, f, 1);
  if (s[i] && !x[i])
    if (carry) ret += solve(i + 1, 0, f, se) + solve(i + 1, 1, 1, 1);
  if (!s[i] && x[i])
    if (carry) ret += solve(i + 1, 1, 1, se) + solve(i + 1, 1, f, 1);
  if (!s[i] && !x[i])
    if (!carry) ret += solve(i + 1, 1, 1, 1) + solve(i + 1, 0, f, se);
  return ret;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> S >> X;
  if (S < X) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 50; i++) {
    if ((long long)(1ULL << i) > S) break;
    s[i] = (S & (long long)(1ULL << i)) ? 1 : 0;
    x[i] = (X & (long long)(1ULL << i)) ? 1 : 0;
    if (s[i]) len = i;
  }
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0, 0, 0) << endl;
}

```