Educational Codeforces Round 35 B. Two Cakes【枚举/给盘子个数，两份蛋糕块数，最少需要在每个盘子放几块蛋糕保证所有蛋糕块都装下】

B. Two Cakes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

It's New Year's Eve soon, so Ivan decided it's high time he started setting the table. Ivan has bought two cakes and cut them into pieces: the first cake has been cut into a pieces, and the second one — into b pieces.

Ivan knows that there will be n people at the celebration (including himself), so Ivan has set n plates for the cakes. Now he is thinking about how to distribute the cakes between the plates. Ivan wants to do it in such a way that all following conditions are met:

1. Each piece of each cake is put on some plate;
2. Each plate contains at least one piece of cake;
3. No plate contains pieces of both cakes.

To make his guests happy, Ivan wants to distribute the cakes in such a way that the minimum number of pieces on the plate is maximized. Formally, Ivan wants to know the maximum possible number x such that he can distribute the cakes according to the aforementioned conditions, and each plate will contain at least x pieces of cake.

Help Ivan to calculate this number x!

Input

The first line contains three integers n, a and b (1 ≤ a, b ≤ 100, 2 ≤ n ≤ a + b) — the number of plates, the number of pieces of the first cake, and the number of pieces of the second cake, respectively.

Output

Print the maximum possible number x such that Ivan can distribute the cake in such a way that each plate will contain at least x pieces of cake.

Examples

input

5 2 3

output

1

input

4 7 10

output

3

Note

In the first example there is only one way to distribute cakes to plates, all of them will have 1 cake on it.

In the second example you can have two plates with 3 and 4 pieces of the first cake and two plates both with 5 pieces of the second cake. Minimal number of pieces is 3.

【题意】：给盘子个数n，两份蛋糕块数a和b，需要在每个盘子最多放几块蛋糕保证所有蛋糕块都装下。

【分析】：从小到大枚举i（放几块蛋糕），若a / i + b / i >= n 则合法，i 的最大枚举量是min(a,b)，不然可能有多余的 a % i 没地方放。

【时间复杂度&&优化】：

【代码】：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,a,b;
    cin>>n>>a>>b;
    for(int i=min(a,b);;i--)
    {
        if(a/i+b/i>=n) return *printf("%d\n",i);
    }
}

技巧枚举