#include <bits/stdc++.h>
const int MAXN = 50;
bool b[MAXN];
long long t[MAXN];
void pre() {
  t[0] = 1;
  for (int i = 1; i <= 42; i++) {
    t[i] = t[i - 1] * 2;
  }
}
int de(long long n) {
  int i = 0;
  while (n) {
    i++;
    n >>= 1;
  }
  return i;
}
int deal(long long m) {
  int ans = 0, i = 0;
  while (m) {
    if (m % 2 == 1) {
      ans++;
      b[i] = true;
    }
    i++;
    m >>= 1;
  }
  return ans;
}
bool judge(long long x, int lenx) {
  for (int i = lenx; i >= 0; i--) {
    if (!b[i]) {
      if (x >= t[i + 1]) {
        x -= t[i + 1];
      }
    }
  }
  if (!x) {
    return false;
  } else {
    return true;
  }
}
int main() {
  long long n, m;
  scanf("%I64d%I64d", &n, &m);
  if (!m) {
    if (n % 2 == 1) {
      printf("0\n");
    } else {
      printf("1\n");
    }
  } else if (n < m) {
    printf("0\n");
  } else {
    pre();
    memset(b, false, sizeof(b));
    int ans = deal(m);
    if (judge(n - m, de(n))) {
      printf("0\n");
    } else {
      if (n == m) {
        printf("%I64d\n", t[ans] - 2);
      } else {
        printf("%I64d\n", t[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>
const int MAXN = 50;
bool b[MAXN];
long long t[MAXN];
void pre() {
  t[0] = 1;
  for (int i = 1; i <= 42; i++) {
    t[i] = t[i - 1] * 2;
  }
}
int de(long long n) {
  int i = 0;
  while (n) {
    i++;
    n >>= 1;
  }
  return i;
}
int deal(long long m) {
  int ans = 0, i = 0;
  while (m) {
    if (m % 2 == 1) {
      ans++;
      b[i] = true;
    }
    i++;
    m >>= 1;
  }
  return ans;
}
bool judge(long long x, int lenx) {
  for (int i = lenx; i >= 0; i--) {
    if (!b[i]) {
      if (x >= t[i + 1]) {
        x -= t[i + 1];
      }
    }
  }
  if (!x) {
    return false;
  } else {
    return true;
  }
}
int main() {
  long long n, m;
  scanf("%I64d%I64d", &n, &m);
  if (!m) {
    if (n % 2 == 1) {
      printf("0\n");
    } else {
      printf("1\n");
    }
  } else if (n < m) {
    printf("0\n");
  } else {
    pre();
    memset(b, false, sizeof(b));
    int ans = deal(m);
    if (judge(n - m, de(n))) {
      printf("0\n");
    } else {
      if (n == m) {
        printf("%I64d\n", t[ans] - 2);
      } else {
        printf("%I64d\n", t[ans]);
      }
    }
  }
  return 0;
}

```