质因数的思维题

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 < X_i - 1 and then finds the minimum X_i ≥ X_i - 1 such that p divides X_i. Note that if the selected prime p already divides X_i - 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

14

Output

6

Input

Copy

20

Output

15

Input

Copy

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.

题目分析 ： 给你一个X2， 让你去寻找一个X0， 并且要求 x0 是最小的，选取的规则是每次选取一个质因数，在变换到X的时候是让其扩大一定的倍数，使其刚好大于等于当前的x

思路分析 ：对于我输入的一个X2，我们发现当它的质因数越大时，上一个 x1 可以选取的值就可以越小，因此预处理下每个数最大的质因数就可以了

代码示例 ：

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int f[maxn];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x;
    cin >> x;
    for(int i = 2; i <= 1000000; i++){
        if (!f[i]){
            for(int j = i+i; j <= 1000000; j += i){
                f[j] = i;
            }
        }
        f[i] = i - f[i] + 1;
    }
    //for(int i = 1; i <= x; i++) printf("%d ", f[i]);
    int ans = inf;
    //cout << f[x] << endl;
    for(int i = f[x]; i <= x; i++){
        ans = min(ans, f[i]);
    }
    cout << ans << endl;
    return 0;
}