codeforces 588 A Duff and Meat

A. Duff and Meat

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 aikilograms of meat.

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.

Sample test(s)

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.

#include <iostream>  
#include <cstdio>  
#include <climits>  
#include <cstring>  
#include <cstdlib>  
#include <cmath>  
#include <vector>  
#include <queue>  
#include <algorithm>  
#define esp 1e-6  
#define inf 0x0f0f0f0f  
#define LL long long    
using namespace std;  
  
/************************************************  
  
Desiner:hl  
time:2016/02/06
Exe.Time:31 ms
Exe.Memory:0 KB  
 
題目鏈接：http://codeforces.com/contest/588/problem/A
 
題意：第i天吃ai kg的食物，每1kg花費pi元，可以存食物，問總共消費是多少 
 
 
題解：簡單貪心 
************************************************/   

int main(){
	int i,j,k,m,n,imin=inf;
	int ai,pi;
	int ans=0;
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d%d",&ai,&pi);
		imin=imin>pi?pi:imin;
		ans=ans+ai*imin;
	}
	printf("%d\n",ans);
}