#include <bits/stdc++.h>
using namespace std;
long long sum, Xor;
long long dp[65][2][2][2];
long long solve(int indx, int carry, bool x, bool y, long long xx = 0,
                long long yy = 0) {
  if (indx == 63) {
    return (x && y && carry == 0);
  }
  long long &res = dp[indx][carry][x][y];
  if (res != -1) return res;
  res = 0;
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < 2; ++j) {
      int a = ((Xor >> indx) & 1), b = ((sum >> indx) & 1);
      if ((i ^ j) == a && (i + j + carry) % 2 == b) {
        long long _x = xx, _y = yy;
        if (i) _x |= (1 << indx);
        if (j) _y |= (1 << indx);
        res += solve(indx + 1, (i + j + carry) / 2, x | i, y | j, _x, _y);
      }
    }
  }
  return res;
}
int main() {
  cin >> sum >> Xor;
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0, 0, 0) << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long sum, Xor;
long long dp[65][2][2][2];
long long solve(int indx, int carry, bool x, bool y, long long xx = 0,
                long long yy = 0) {
  if (indx == 63) {
    return (x && y && carry == 0);
  }
  long long &res = dp[indx][carry][x][y];
  if (res != -1) return res;
  res = 0;
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < 2; ++j) {
      int a = ((Xor >> indx) & 1), b = ((sum >> indx) & 1);
      if ((i ^ j) == a && (i + j + carry) % 2 == b) {
        long long _x = xx, _y = yy;
        if (i) _x |= (1 << indx);
        if (j) _y |= (1 << indx);
        res += solve(indx + 1, (i + j + carry) / 2, x | i, y | j, _x, _y);
      }
    }
  }
  return res;
}
int main() {
  cin >> sum >> Xor;
  memset(dp, -1, sizeof dp);
  cout << solve(0, 0, 0, 0) << endl;
  return 0;
}

```