#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
map<int, int> mp;
int bitx[60], bity[60];
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    if ((s - x) % 2 || s < x) {
      printf("0\n");
      continue;
    }
    long long y = (s - x) / 2;
    int ctx = 0, cty = 0;
    long long ty = y;
    memset(bitx, 0, sizeof(bitx));
    memset(bity, 0, sizeof(bity));
    while (x) {
      bitx[++ctx] = x % 2;
      x /= 2;
    }
    while (y) {
      bity[++cty] = y % 2;
      y /= 2;
    }
    long long ans = 1;
    for (int i = 1; i <= max(ctx, cty); i++) {
      if (bitx[i] == 1 && bity[i] == 1) ans = 0;
      if (bitx[i] == 0 && bity[i] == 0) ans *= 1;
      if (bitx[i] == 0 && bity[i] == 1) ans *= 1;
      if (bitx[i] == 1 && bity[i] == 0) ans *= 2;
    }
    if (ty == 0) ans -= 2;
    printf("%I64d\n", ans);
  }
  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;
const int N = 1005;
map<int, int> mp;
int bitx[60], bity[60];
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    if ((s - x) % 2 || s < x) {
      printf("0\n");
      continue;
    }
    long long y = (s - x) / 2;
    int ctx = 0, cty = 0;
    long long ty = y;
    memset(bitx, 0, sizeof(bitx));
    memset(bity, 0, sizeof(bity));
    while (x) {
      bitx[++ctx] = x % 2;
      x /= 2;
    }
    while (y) {
      bity[++cty] = y % 2;
      y /= 2;
    }
    long long ans = 1;
    for (int i = 1; i <= max(ctx, cty); i++) {
      if (bitx[i] == 1 && bity[i] == 1) ans = 0;
      if (bitx[i] == 0 && bity[i] == 0) ans *= 1;
      if (bitx[i] == 0 && bity[i] == 1) ans *= 1;
      if (bitx[i] == 1 && bity[i] == 0) ans *= 2;
    }
    if (ty == 0) ans -= 2;
    printf("%I64d\n", ans);
  }
  return 0;
}

```