Codeforces Round #554 (Div. 2) C. Neko does Maths

C. Neko does Maths

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Neko has two integers aa and bb. His goal is to find a non-negative integer kk such that the least common multiple of a+ka+k and b+kb+k is the smallest possible. If there are multiple optimal integers kk, 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 aa and bb (1≤a,b≤1091≤a,b≤109).

Output

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

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

Examples

input

Copy

6 10

output

Copy

2

input

Copy

21 31

output

Copy

9

input

Copy

5 10

output

Copy

0

Note

In the first test, one should choose k=2k=2, as the least common multiple of 6+26+2 and 10+210+2 is 2424, which is the smallest least common multiple possible.

首先有个结论 gcd(a,b)=gcd(a,b-a), 因为假设gcd(a,b)=c,那么a%c=b%c=0,又有(a-b)%c=0,所以 gcd(a,b)=gcd(a,b-a),根据题意，让lcm最小那么就是要求最大的gcd，最大的gcd必定是两数中最大的约数，由于题目中b-a是定值，所以就可以枚举b-a的约数，然后把a凑到含有此约数为止，因此需要凑的值k=x-a%x，当然，a%x=0时k=0.除此之外记得开ll。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=1e5+5;
ll n,m,a,b;
vector<int> v;

ll gcd(ll x,ll y)
{
    return y==0 ? x: gcd(y,x%y);
}

ll lcm(ll x,ll y)
{
    return x*y/gcd(x,y);
}


int main()
{
    cin>>a>>b;
    if(a>b)
    {
        swap(a,b);
    }
    ll s=b-a;
    for(int i=1; i*i<=s; i++)
    {
        if(s%i==0)
        {
            v.push_back(i);
            if(i*i!=s)
                v.push_back(s/i);
        }
    }
    ll ans=1e18+5,k;
    for(int i=0; i<v.size(); i++)
    {
        ll t=0,x=v[i];
        if(a%x!=0)
        {
            t=x-a%x;
        }
        ll temp=lcm(a+t,b+t);
        if(ans>temp)
        {
            ans=temp;
            k=t;
        }
    }
    if(a==b) k=0;
    cout<<k<<endl;
    return 0;
}