CF408A.Line to Cashier

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 k__i people. The j-th person standing in the queue to the i-th cashier has m__i, 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.

小瓦西娅去超市购买一些杂货。他在超市里逛了很长时间，买了一篮子商品。现在他需要选择一个收银台来付款。

超市出口处共有 $n$ 个收银台。当前第 $i$ 个收银台前的队伍中已有 $k_i$ 人。排在第 $i$ 个收银台队伍中第 $j$ 位的顾客，其购物篮中有 $m_{i,j}$ 件商品。瓦西娅知道：

• 收银员扫描一件商品需要 5 秒；
• 收银员扫描完某位顾客的所有商品后，还需额外花费 15 秒来收取该顾客的款项并找零。

当然，瓦西娅希望选择一个队列，以便能尽快离开超市。请帮他编写一个程序，输出瓦西娅到达任一收银台所需的最短时间（单位：秒）。

The first line contains integer n (1 ≤ n ≤ 100) — the number of cashes in the shop. The second line contains n space-separated integers: _k_1, _k_2, ..., k__n (1 ≤ k__i ≤ 100), where k__i is the number of people in the queue to the i-th cashier.

The i-th of the next n lines contains k__i space-separated integers: m__i, 1, m__i, 2, ..., m__i, k__i (1 ≤ m__i, j ≤ 100) — the number of products the j-th person in the queue for the i-th cash has.

第一行包含一个整数 $n$（$1 \leq n \leq 100$）—— 商店中收银台的数量。
第二行包含 $n$ 个用空格分隔的整数：$k_1,\,k_2,\,\dots,\,k_n$（$1 \leq k_i \leq 100$），其中 $k_i$ 表示第 $i$ 个收银台前排队的人数。

接下来的 $n$ 行中，第 $i$ 行包含 $k_i$ 个用空格分隔的整数：$m_{i,1},\,m_{i,2},\,\dots,\,m_{i,k_i}$（$1 \leq m_{i,j} \leq 100$）—— 表示第 $i$ 个收银台队列中第 $j$ 个人所购商品的数量。

Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.

输出一个整数——Vasya 到达收银台所需的最少秒数。

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.

在第二个测试样例中，如果瓦西娅选择第一个队列，他将在 $100\cdot5\,+\,15\,=\,515$ 秒后到达收银台。但如果他选择第二个队列，则需要 $1\cdot5\,+\,2\cdot5\,+\,2\cdot5\,+\,3\cdot5\,+\,4\cdot15\,=\,100$ 秒；选择第三个队列则需要 $1\cdot5\,+\,9\cdot5\,+\,1\cdot5\,+\,3\cdot15\,=\,100$ 秒；而选择第四个队列则需要 $7\cdot5\,+\,8\cdot5\,+\,2\cdot15\,=\,105$ 秒。因此，瓦西娅选择第二个或第三个队列时能更快到达收银台。