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 a_i 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_i dollars per kilogram. Malek knows all numbers a₁, ..., a_n and p₁, ..., p_n. 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 a_i and p_i (1 ≤ a_i, p_i ≤ 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 <string>
#include <cstring>
#include <algorithm>
using namespace std;

int a[100010];
int p[100010];

int main() {
	int n;
	cin >> n;
	memset(a, 0, sizeof(a));
	for (int i = 0; i<n; i++)
		cin >> a[i]>> p[i];
	int res = 0, min = p[0];
	for (int i = 0; i<n; i++) {
		if (p[i] < min)
			min = p[i];
		res += min*a[i];
	}
	cout << res<< endl;
	return 0;
}