codeforces_665B. Shopping

B. Shopping

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Example

input

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

output

14

Note

Customer 1 wants the items 1 and 5.

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

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

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

Customer 2 wants the items 3 and 1.

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

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

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

Total time is 8 + 6 = 14.

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

有一段时间没做题了，看题看了老半天，水题一道，暴力即可。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

int a[110];
int n,m,k,ans=0;

void change(int x)
{
int pos=0;
for(int i=1;i<=k;i++)
{
if(a[i]==x)
{
pos=i;break;
}
}
ans+=pos;
for(int i=pos;i>1;i--)
{
a[i]=a[i-1];
}
a[1]=x;
}

int main()
{
int i,j;
Si(n),Si(m),Si(k);
for(i=1;i<=k;i++)
{
Si(a[i]);
}
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
int x;
Si(x);
change(x);
}
}
Pi(ans);
return 0;
}