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.

首先有個結論 gcd(a,b)=gcd(a,b-a), 因為假設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),根據題意，讓lcm最小那么就是要求最大的gcd，最大的gcd必定是兩數中最大的約數，由於題目中b-a是定值，所以就可以枚舉b-a的約數，然后把a湊到含有此約數為止，因此需要湊的值k=x-a%x，當然，a%x=0時k=0.除此之外記得開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;
}