CodeForces - 408A A. Line to Cashier

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.

Examples

input

1
1
1

output

20

input

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

output

100

题意：有n名收银员，每个收银员前有s[i]个人，每个人手里有k个东西，收银员扫描一个物体花5秒，给人找零用15秒，找出所有收银员结算完用时最短的。

AC代码：

#include<cstdio>
#include<algorithm>

using namespace std;

int main(){
	int n;
	int s[110],s1;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&s[i]);
	}
	int sum=0,minn=99999999;
	for(int i=1;i<=n;i++){
		sum=0;
		for(int j=1;j<=s[i];j++){
			scanf("%d",&s1);
			sum+=5*s1;
		}
		sum+=15*s[i];
		minn=min(sum,minn);
	}
	printf("%d",minn);
	return 0;
}