#include <bits/stdc++.h>
long long read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
  return res * flag;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
long long s, x;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long count(long long x) {
  long long res = 0;
  while (x) res += x & 1, x >>= 1;
  return res;
}
void solution() {
  s = read(), x = read();
  42;
  print((s < x || ((s - x) & (x << 1 | 1)))
            ? 0
            : qpow(2, count(x)) - ((s == x) * 2)),
      puts("");
}
signed main() {
  42;
  solution();
}

### Prompt

Generate 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>
long long read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
  return res * flag;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
long long s, x;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long count(long long x) {
  long long res = 0;
  while (x) res += x & 1, x >>= 1;
  return res;
}
void solution() {
  s = read(), x = read();
  42;
  print((s < x || ((s - x) & (x << 1 | 1)))
            ? 0
            : qpow(2, count(x)) - ((s == x) * 2)),
      puts("");
}
signed main() {
  42;
  solution();
}

```