VK Cup 2018 Round 1: A. Primal Sport

A. Primal Sport

time limit per test 1.5 seconds

memory limit per test 256 megabytes

input standard input

output standard output

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 palready 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

input

Copy

output

input

Copy

output

input

Copy

output

题意：初始有一个x0，你可以选择任意一个p满足p<x0且p是质数，之后计算x1为≥x0最小的p的倍数

然后再通过x1按照一样的算法得出一个x2

现给你一个x2，问x0最小可能是多少，保证有解（x2绝对不是质数）

稍加思考会发现，对于当前x2，x1的可能取值范围为[x2-p+1, x2]，其中p为x2的最大质因子

暴力[x2-p+1, x2]中的每一个数作为x1，如果x1是质数就无视（因为这样就不可能存在x0），否则计算x0 = x1-p'+1

其中p'是x1的最大质因子，很显然最后答案就是min(x0)

#include<stdio.h>
#include<algorithm>
using namespace std;
int cnt, flag[1005] = {1,1}, pri[1005];
int Cha(int x)
{int i, bet;bet = 0;for(i=1;pri[i]*pri[i]<=x&&pri[i]!=0;i++){while(x%pri[i]==0){bet = max(bet, pri[i]);x /= pri[i];}}if(x!=1)bet = max(bet, x);return bet;
}
int main(void)
{int i, j, x2, now, ans;for(i=2;i<=1000;i++){if(flag[i])continue;pri[++cnt] = i;for(j=i*i;j<=1000;j+=i)flag[j] = 1;}scanf("%d", &x2);now = Cha(x2);ans = 10000000;for(i=x2-now+1;i<=x2;i++){now = Cha(i);if(now==i)continue;ans = min(ans, i-now+1);}printf("%d\n", ans);return 0;
}

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