Codeforce 588A - Duff and Meat （贪心）

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.

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.

题解：不断更新肉价的最小值

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cstdlib>
 #include <iomanip>
 #include <cmath>
 #include <ctime>
 #include <map>
 #include <set>
 using namespace std;
 #define lowbit(x) (x&(-x))
 #define max(x,y) (x>y?x:y)
 #define min(x,y) (x<y?x:y)
 #define MAX 100000000000000000
 #define MOD 1000000007
 #define pi acos(-1.0)
 #define ei exp(1)
 #define PI 3.141592653589793238462
 #define INF 0x3f3f3f3f3f
 #define mem(a) (memset(a,0,sizeof(a)))
 typedef long long ll;
 const int N=;
 const int mod=1e9+;
 int a[N],b[N];
 int main()
 {
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++){
         scanf("%d %d",&a[i],&b[i]);
     }
     ll s=;
     s+=a[]*b[];
     int t=b[];
     for(int i=;i<n;i++){
         if(b[i]<t) t=b[i];
         s+=a[i]*t;
     }
     printf("%lld\n",s);
     return ;
 }