A. Line to Cashier

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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

Sample test(s)

input

1
1
1

output

20

input

4
1 4 3 2
100
1 2 2 3
1 9 1
7 8

output

100

Note

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.

解题说明：此题是一道简单的模拟题，统计每一行数字对应花费的时间，找到一个最小值即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
using namespace std;

int main()
{
	int n, i, j;
	int a[101], f;
	long min, k;
	scanf("%d", &n);
	min= 1000000;
	for (i = 0; i<n; i++)
	{
		scanf("%d", &a[i]);
	}
	for (i = 0; i<n; i++)
	{
		k = 0;
		for (j = 0; j<a[i]; j++)
		{
			scanf("%d", &f);
			k += (f * 5 + 15);

		}
		if (k<min)
		{
			min = k;
		}
	}
	printf("%ld\n", min);
	return 0;
}