#include <bits/stdc++.h>
using namespace std;
long long dp[105][4][2];
long long s, x;
int main() {
  while (scanf("%lld%lld", &s, &x) != EOF) {
    if (s < x) {
      printf("0\n");
      continue;
    }
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    int loc;
    long long ans = 0;
    if (s == x) ans = -2;
    for (int i = 1; s; i++) {
      int tx = x % 2, ts = s % 2;
      if (tx == 1) {
        if (ts == 1) {
          for (int j = 0; j <= 3; j++) {
            dp[i][1][0] += dp[i - 1][j][0];
            dp[i][2][0] += dp[i - 1][j][0];
          }
        } else {
          for (int j = 0; j <= 3; j++) {
            dp[i][1][1] += dp[i - 1][j][1];
            dp[i][2][1] += dp[i - 1][j][1];
          }
        }
      } else {
        if (ts == 1) {
          for (int j = 0; j <= 3; j++) {
            dp[i][0][0] += dp[i - 1][j][1];
            dp[i][3][1] += dp[i - 1][j][1];
          }
        } else {
          for (int j = 0; j <= 3; j++) {
            dp[i][0][0] += dp[i - 1][j][0];
            dp[i][3][1] += dp[i - 1][j][0];
          }
        }
      }
      loc = i;
      s /= 2;
      x /= 2;
    }
    for (int i = 0; i <= 3; i++) {
      ans += dp[loc][i][0];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 dp[105][4][2];
long long s, x;
int main() {
  while (scanf("%lld%lld", &s, &x) != EOF) {
    if (s < x) {
      printf("0\n");
      continue;
    }
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    int loc;
    long long ans = 0;
    if (s == x) ans = -2;
    for (int i = 1; s; i++) {
      int tx = x % 2, ts = s % 2;
      if (tx == 1) {
        if (ts == 1) {
          for (int j = 0; j <= 3; j++) {
            dp[i][1][0] += dp[i - 1][j][0];
            dp[i][2][0] += dp[i - 1][j][0];
          }
        } else {
          for (int j = 0; j <= 3; j++) {
            dp[i][1][1] += dp[i - 1][j][1];
            dp[i][2][1] += dp[i - 1][j][1];
          }
        }
      } else {
        if (ts == 1) {
          for (int j = 0; j <= 3; j++) {
            dp[i][0][0] += dp[i - 1][j][1];
            dp[i][3][1] += dp[i - 1][j][1];
          }
        } else {
          for (int j = 0; j <= 3; j++) {
            dp[i][0][0] += dp[i - 1][j][0];
            dp[i][3][1] += dp[i - 1][j][0];
          }
        }
      }
      loc = i;
      s /= 2;
      x /= 2;
    }
    for (int i = 0; i <= 3; i++) {
      ans += dp[loc][i][0];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```