CF665B.Shopping

普及/提高-

通过率:0%

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题目描述

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 kk items. nn customers have already used the above service. Each user paid for mm items. Let aija_{ij} denote the jj -th item in the ii -th person's order.

Due to the space limitations all the items are arranged in one single row. When Ayush receives the ii -th order he will find one by one all the items aija_{ij} ( 1<=j<=m1<=j<=m ) in the row. Let pos(x)pos(x) denote the position of the item xx in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1)+pos(ai2)+...+pos(aim)pos(a_{i1})+pos(a_{i2})+...+pos(a_{im}) for the ii -th customer.

When Ayush accesses the xx -th element he keeps a new stock in the front of the row and takes away the xx -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.

输入格式

The first line contains three integers nn , mm and kk ( 1<=n,k<=100,1<=m<=k1<=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 kk distinct integers plp_{l} ( 1<=pl<=k1<=p_{l}<=k ) denoting the initial positions of the items in the store. The items are numbered with integers from 11 to kk .

Each of the next nn lines contains mm distinct integers aija_{ij} ( 1<=aij<=k1<=a_{ij}<=k ) — the order of the ii -th person.

输出格式

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

输入输出样例

  • 输入#1

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

    输出#1

    14
    

说明/提示

Customer 11 wants the items 11 and 55 .

pos(1)=3pos(1)=3 , so the new positions are: [1,3,4,2,5][1,3,4,2,5] .

pos(5)=5pos(5)=5 , so the new positions are: [5,1,3,4,2][5,1,3,4,2] .

Time taken for the first customer is 3+5=83+5=8 .

Customer 22 wants the items 33 and 11 .

pos(3)=3pos(3)=3 , so the new positions are: [3,5,1,4,2][3,5,1,4,2] .

pos(1)=3pos(1)=3 , so the new positions are: [1,3,5,4,2][1,3,5,4,2] .

Time taken for the second customer is 3+3=63+3=6 .

Total time is 8+6=148+6=14 .

Formally pos(x)pos(x) is the index of xx in the current row.

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