Primal Sport

Alice and Bob begin their day with a quick game. They first choose a starting number X0 ≥ 3 and try to reach one million by the process described below.

Alice goes first and then they take alternating turns. In the i-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.

Formally, he or she selects a prime p < Xi - 1 and then finds the minimum Xi ≥ Xi - 1 such that p divides Xi. Note that if the selected prime p already divides Xi - 1, then the number does not change.

Eve has witnessed the state of the game after two turns. Given X2, help her determine what is the smallest possible starting number X0. Note that the players don’t necessarily play optimally. You should consider all possible game evolutions.

Input
The input contains a single integer X2 (4 ≤ X2 ≤ 106). It is guaranteed that the integer X2 is composite, that is, is not prime.

Output
Output a single integer — the minimum possible X0.

Examples
inputCopy
14
output
6
inputCopy
20
output
15
inputCopy
8192
output
8191
Note
In the first test, the smallest possible starting number is X0 = 6. One possible course of the game is as follows:

Alice picks prime 5 and announces X1 = 10
Bob picks prime 7 and announces X2 = 14.
In the second case, let X0 = 15.

Alice picks prime 2 and announces X1 = 16
Bob picks prime 5 and announces X2 = 20.
思路：设find[x]表示x的最大质因子，由题意可知，x1范围在[x2-find[x2]+1,x2],x0范围在[x1-find[x1]+1,x1]枚举x1，比较不同x1得到的最小x0的最小值。

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
int vis[1000005],Find[1000005];
int x2,x0;
void getprime()
{
    for(int i=2;i<=1000005;i++)
    {       
        if(!vis[i])
        {
            Find[i]=-1;
            for(int j=i*2;j<=1000005;j+=i)
            {
                vis[j]=1;
                Find[j]=i;
            }
        }
    }
} 
int main()
{
        getprime();
        scanf("%d",&x2);
        x0=1000005;
        for(int x1=x2-Find[x2]+1;x1<=x2;x1++)
        {
            x0=min(x0,x1-Find[x1]+1);
        }
        printf("%d\n",x0);
        return 0;

}