CF408A.Line to Cashier
普及/提高-
通过率:0%
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题目描述
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are n cashiers at the exit from the supermarket. At the moment the queue for the i -th cashier already has ki people. The j -th person standing in the queue to the i -th cashier has mi,j items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item;
- after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
输入格式
The first line contains integer n ( 1<=n<=100 ) — the number of cashes in the shop. The second line contains n space-separated integers: k1,k2,...,kn ( 1<=ki<=100 ), where ki is the number of people in the queue to the i -th cashier.
The i -th of the next n lines contains ki space-separated integers: mi,1,mi,2,...,mi,ki ( 1<=mi,j<=100 ) — the number of products the j -th person in the queue for the i -th cash has.
输出格式
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
输入输出样例
输入#1
1 1 1
输出#1
20
输入#2
4 1 4 3 2 100 1 2 2 3 1 9 1 7 8
输出#2
100
说明/提示
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100⋅5+15=515 seconds. But if he chooses the second queue, he will need 1⋅5+2⋅5+2⋅5+3⋅5+4⋅15=100 seconds. He will need 1⋅5+9⋅5+1⋅5+3⋅15=100 seconds for the third one and 7⋅5+8⋅5+2⋅15=105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.