Supermarket

We often go to supermarkets to buy some fruits or vegetables, and on the tag there prints the price for a kilo. But in some supermarkets, when asked how much the items are, the clerk will say that a yuan for b kilos (You don’t need to care about what “yuan” is), the same as a/b yuan for a kilo.

Now imagine you’d like to buy m kilos of apples. You’ve asked n supermarkets and got the prices. Find the minimum cost for those apples.

You can assume that there are enough apples in all supermarkets.v

Input
The first line contains two positive integers n and m (1≤n≤5000, 1≤m≤100), denoting that there are n supermarkets and you want to buy m kilos of apples.

The following n lines describe the information of the supermarkets. Each line contains two positive integers a,b (1≤a,b≤100), denoting that in this supermarket, you are supposed to pay a yuan for b kilos of apples.

Output
The only line, denoting the minimum cost for m kilos of apples. Please make sure that the absolute or relative error between your answer and the correct answer won’t exceed 10−6.

Formally, let your answer be x, and the jury’s answer be y. Your answer is considered correct if |x−y|max(1,|y|)≤10−6.

Examples

Input
3 5
1 2
3 4
1 3

Output
1.66666667

Input
2 1
99 100
98 99

Output
0.98989899

Note
In the first sample, you are supposed to buy 5 kilos of apples in supermarket 3. The cost is 5/3 yuan.

In the second sample, you are supposed to buy 1 kilo of apples in supermarket 2. The cost is 98/99 yuan.

代码略显复杂，思路很简单：算出每1kg水果的单价，将第一个超市给出的价格设为min，不断输入下一超市给出的价格w，一旦遇到价格小于第一个超市给出的价格，将新的价格赋值给min，最终算出欲买mkg水果所需的总价s。

/*by BFU zq
  2021/1/15 15:31:01*/
#include<bits/stdc++.h>
using namespace std;

int main()
{
	int m,n,i=0,ans=0;
	double a,b,w,s,min;
	
	cin>>n>>m;
	cin>>a>>b;
	min = a / b;
	i++;
	
	for (i=1;i<n;i++)
	{
		cin>>a>>b
		w = a / b;
		
		if (w < min)
		{
			min = w;
		}
		else ans++;
	}
	
	s = min * m;
	cout<<s<<endl;
	
	return 0;
}