#include <bits/stdc++.h>
using namespace std;
int wei[60], pos[60];
map<int, int> mp;
map<int, int>::iterator it;
int main() {
  long long s, x;
  int i, j, len = 0;
  while (scanf("%I64d%I64d", &s, &x) != EOF) {
    long long t = x;
    while (t) {
      wei[len++] = t % 2;
      t /= 2;
    }
    mp.clear();
    int num = 0;
    for (i = 0; i < len; i++) {
      if (wei[i] == 1) {
        num++;
        s -= (1LL << i);
      } else {
        mp[i]++;
      }
    }
    for (i = len; i <= 50; i++) {
      mp[i]++;
    }
    if (s < 0) {
      printf("0\n");
      continue;
    }
    int tot = 0;
    int flag = 1;
    long long ss = s;
    while (ss) {
      if (ss % 2 == 1) {
        if (tot == 0 || mp[tot - 1] == 0) {
          flag = 0;
          break;
        }
      }
      tot++;
      ss /= 2;
    }
    if (flag) {
      if (s == 0)
        printf("%I64d\n", (1LL << num) - 2);
      else
        printf("%I64d\n", (1LL << num));
    } else
      printf("0\n");
  }
  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;
int wei[60], pos[60];
map<int, int> mp;
map<int, int>::iterator it;
int main() {
  long long s, x;
  int i, j, len = 0;
  while (scanf("%I64d%I64d", &s, &x) != EOF) {
    long long t = x;
    while (t) {
      wei[len++] = t % 2;
      t /= 2;
    }
    mp.clear();
    int num = 0;
    for (i = 0; i < len; i++) {
      if (wei[i] == 1) {
        num++;
        s -= (1LL << i);
      } else {
        mp[i]++;
      }
    }
    for (i = len; i <= 50; i++) {
      mp[i]++;
    }
    if (s < 0) {
      printf("0\n");
      continue;
    }
    int tot = 0;
    int flag = 1;
    long long ss = s;
    while (ss) {
      if (ss % 2 == 1) {
        if (tot == 0 || mp[tot - 1] == 0) {
          flag = 0;
          break;
        }
      }
      tot++;
      ss /= 2;
    }
    if (flag) {
      if (s == 0)
        printf("%I64d\n", (1LL << num) - 2);
      else
        printf("%I64d\n", (1LL << num));
    } else
      printf("0\n");
  }
  return 0;
}

```