题目连接：

http://www.codeforces.com/contest/665/problem/B

Description

Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.

The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.

Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.

When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.

Your task is to calculate the total time it takes for Ayush to process all the orders.

You can assume that the market has endless stock.

Input

The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.

The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.

Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.

Output

Print the only integer t — the total time needed for Ayush to process all the orders.

Sample Input

2 2 5
3 4 1 2 5
1 5
3 1

Sample Output

14

Hint

题意

现在有k件商品，每个商品的位置已经告诉你了

现在有n个人，每个人有m个需求，每个需求就是要把第a[i][j]个物品拿到第一个位置来

他的代价是pos[a[i][j]]

问你所有代价是多少

题解：

数据范围太小了，所以直接暴力就好了~

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 205;
int a[maxn],b[maxn],t=0;
int main()
{int ans = 0;int n,m,k;scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=k;i++)scanf("%d",&a[i]);for(int i=1;i<=n*m;i++){int x;scanf("%d",&x);int pos = 0;for(int j=1;j<=k;j++)if(x==a[j])pos=j;ans+=pos;b[1]=a[pos];t=2;for(int j=1;j<=k;j++)if(j!=pos)b[t++]=a[j];for(int j=1;j<=k;j++)a[j]=b[j];}cout<<ans<<endl;
}