Codeforces Round #326 (Div. 2) A. Duff and Meat

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本文介绍了一个关于肉价波动的问题，通过最优购买策略来最小化Duff在n天内食用肉品的总花费。Duff每天需要吃特定数量的肉，而肉的价格每天都在变化，可以提前购买储备。文章提供了具体的实现代码。

Duff is addicted to meat! Malek wants to keep her happy for n days. In order to be happy in i-th day, she needs to eat exactly ai kilograms of meat.

$\text{[math]}$

There is a big shop uptown and Malek wants to buy meat for her from there. In i-th day, they sell meat for pi dollars per kilogram. Malek knows all numbers a1, ..., an and p1, ..., pn. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.

Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for n days.

Input

The first line of input contains integer n (1 ≤ n ≤ 105), the number of days.

In the next n lines, i-th line contains two integers ai and pi (1 ≤ ai, pi ≤ 100), the amount of meat Duff needs and the cost of meat in that day.

Output

Print the minimum money needed to keep Duff happy for n days, in one line.

Examples

input
3
1 3
2 2
3 1

output
10

input
3
1 3
2 1
3 2

output
8

Note

In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.

In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.

题意：

Duff 喜欢吃肉，肉每天的价格为ai，而Duff每天要吃pi斤肉，问n天Duff最少要吃多少钱的肉，肉是可以提前买的（价格低的时候）。

思路：

计算第i天的时候，价格用p1~pi中的最小值计算，不断更新价格的最小值。

#include <iostream>
using namespace std;


int main()
{
    int t;
    int n,m;
    cin>>t;
    int Min=1<<30;
    int sum=0;
    while(t--)
    {
        cin>>n>>m;
        if(m<Min)
        {
            Min=m;
        }
        sum+=n*Min;
    }
    cout<<sum<<endl;
    return 0;
}