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 *a*<sub class="lower-index">*i*</sub> 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 *p*<sub class="lower-index">*i*</sub> dollars per kilogram. Malek knows all numbers *a*<sub class="lower-index">1</sub>, ..., *a*<sub class="lower-index">*n*</sub> and *p*<sub class="lower-index">1</sub>, ..., *p*<sub class="lower-index">*n*</sub>. 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.
The first line of input contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>), the number of days.

In the next *n* lines, *i*-th line contains two integers *a*<sub class="lower-index">*i*</sub> and *p*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub>, *p*<sub class="lower-index">*i*</sub> ≤ 100), the amount of meat Duff needs and the cost of meat in that day.
#include<bits/stdc++.h>
using namespace std;
 
#define cy cout<<"YES"<<endl;
#define cn cout<<"NO"<<endl;
#define cend cout<<endl;
#define ll long long
#define modulo 10^9+7
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    int a[t],p[t];
    cin >> a[0] >> p[0];
    int count = a[0]*p[0],min = p[0];
    for(int i=1;i<t;i++){
        cin >> a[i] >> p[i];
        if(min < p[i]){
            count += min*a[i];
        }else{
            min = p[i];
            count += min*a[i];
        }
    }
    cout << count;
 
return 0;
}