Codeforces Round #467 (Div. 2) B. Vile Grasshoppers[求去掉2-y中所有2-p的数的倍数后剩下的最大值]

B. Vile Grasshoppers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The weather is fine today and hence it's high time to climb the nearby pine and enjoy the landscape.

The pine's trunk includes several branches, located one above another and numbered from 2 to y. Some of them (more precise, from 2 to p) are occupied by tiny vile grasshoppers which you're at war with. These grasshoppers are known for their awesome jumping skills: the grasshopper at branch x can jump to branches .

Keeping this in mind, you wisely decided to choose such a branch that none of the grasshoppers could interrupt you. At the same time you wanna settle as high as possible since the view from up there is simply breathtaking.

In other words, your goal is to find the highest branch that cannot be reached by any of the grasshoppers or report that it's impossible.

Input

The only line contains two integers p and y (2 ≤ p ≤ y ≤ 10⁹).

Output

Output the number of the highest suitable branch. If there are none, print -1 instead.

Examples

Input

Copy

3 6

Output

5

Input

Copy

3 4

Output

-1

Note

In the first sample case grasshopper from branch 2 reaches branches 2, 4 and 6 while branch 3 is initially settled by another grasshopper. Therefore the answer is 5.

It immediately follows that there are no valid branches in second sample case.

[分析]: 给你两个数p，y，求去掉2-y中所有2-p的数的倍数后剩下的最大值，如果没有剩下数输出-1.

[代码]:和筛法一个意思,注意最大值可以逆向枚举.只要算到sqrt（10^9）（i<=p这个条件的加入是为了防止x可能不是素数，但是x不含有2~p之间的任何一个因数）

[代码]:

#include<bits/stdc++.h>
using namespace std;
int p, y;

bool ck(int n) {
    for(int i = ; i * i <= n && i <= p; i++) {
        if(n % i == ) {
            return false;
        }
    }
    return true;
}

int main() {
    cin >> p >> y;
    for(int i = y; i > p; i--) {
        if(ck(i)) {
            printf("%d\n", i);
            return ;
        }
    }
    puts("-1");
    return ;
}

筛法