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.
#creating three empty lists
price=[]
amount=[]
products=[]
#declaring a variable
optimal_price=0
minimum=120
#number of elements as input
n=int(input('dollar:'))
m=int(input("kg:"))
#iterating till the range
for j in range(0,m):
    sle=int(input())
    amount.append(sle)
for i in range(0,n):
    ele=int(input())
    price.append(ele)
#do a loop over price
for idx,item in enumerate(price):
    if(item>minimum):     #making a condition
        price[idx]=minimum
    else:
        minimum=price[idx]
#multiplication between  two lists
for num1,num2 in zip(price,amount):
    products.append(num1*num2)
for gle in range(0,len(products)):
    optimal_price=optimal_price+products[gle]
print(optimal_price)