codeforces. problem/408/A

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 thej-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

111

output

20

input

41 4 3 21001 2 2 31 9 17 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.

#include <string.h>#include <stdio.h>#include <algorithm>#include <iostream>using namespace std;int a[105],b[105][105],sum[105];int main(){    int n;    while(~scanf("%d",&n))    {        memset(sum,0,sizeof(sum));        for(int i=0;i<n;i++)             cin >> a[i];        for(int i=0;i<n;i++)        {            for(int j=0;j<a[i];j++)            {                 cin >> b[i][j];                 sum[i]+=b[i][j]*5;            }            sum[i]+=a[i]*15;        }        sort(sum,sum+n);        printf("%d\n",sum[0]);    }    return 0;}