Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.

Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.

Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?

Input

The only line contains two integers a and b (1 ≤ a, b ≤ 10^9).

Output

Print the smallest non-negative integer k (k ≥ 0) such that the lowest common multiple of a+k and b+k is the smallest possible.

If there are many possible integers k giving the same value of the least common multiple, print the smallest one.

Examples

Input


6 10


Output


2

Input


21 31


Output


9

Input


5 10


Output


0

Note

In the first test, one should choose k = 2, as the least common multiple of 6 + 2 and 10 + 2 is 24, which is the smallest least common multiple possible.
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int x, long long int y) {
  if (x < y) swap(x, y);
  if (x % y == 0) return y;
  return gcd(y, x % y);
}
long long int pwm(long long int a, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      ans = (ans * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    n /= 2;
  }
  return ans % 1000000007;
}
const int pwn = 100010;
long long int pw[pwn];
void pwtwo() {
  pw[0] = 1;
  for (int i = 1; i < pwn; i++) {
    pw[i] = (pw[i - 1] * 2) % 1000000007;
  }
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y);
long long int modInverse(long long int a, long long int m) {
  if (a == 1) return 1;
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  if (g != 1) {
    return -1;
  } else {
    long long int res = (x % m + m) % m;
    return res;
  }
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int fact[1010];
long long int init() {
  fact[0] = 1;
  for (int i = 1; i < 1010; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long int ncr(long long int n, long long int r) {
  long long int ans = fact[n];
  ans = (ans * modInverse(fact[r], 1000000007)) % 1000000007;
  ans = (ans * modInverse(fact[n - r], 1000000007)) % 1000000007;
  return ans;
}
long long int t[42];
bool check(int x) {
  for (int i = 1; i <= 40; i++) {
    if (x == t[i] - 1) return true;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int a, b;
  cin >> a >> b;
  if (a == 1 || b == 1) {
    cout << 0 << endl;
    return 0;
  }
  if (a == b) {
    cout << 0 << endl;
    return 0;
  }
  long long int x = abs(b - a);
  long long int ans = a * b;
  ans = ans / gcd(a, b);
  long long int c = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if ((x % i) == 0) {
      long long int y = a % i;
      y = (i - y) % i;
      long long int k = (a + y) * (b + y);
      k = k / gcd(a + y, b + y);
      if (ans > k) {
        ans = k;
        c = y;
      }
      int z = x / i;
      y = a % z;
      y = (z - y) % z;
      k = (a + y) * (b + y);
      k = k / gcd(a + y, b + y);
      if (ans > k) {
        ans = k;
        c = y;
      }
    }
  }
  cout << c << endl;
  return 0;
}
