Educational Codeforces Round 12（B）暴力模拟

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 aijdenote 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.

题意：题意猜了好久、、、、、、、好吧，有n个顾客，每个顾客要买m件物品，每个物品都有唯一编号，但是这个物品拿出来只能排队拿出来，拿出来后买走了再把这个买走的物品补充到队尾

题解：直接使用双端队列模拟一下就好

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;

#define N int(1e3)
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
typedef long long LL;
deque<int>v;
int main()
{
#ifdef CDZSC
	freopen("i.txt", "r", stdin);
	//freopen("o.txt","w",stdout);
	int _time_jc = clock();
#endif
	int n, m, k,x;
	while (~scanf("%d%d%d",&n,&m,&k))
	{
		v.clear();
		for (int i = 1; i <= k; i++)
		{
			scanf("%d", &x);
			v.push_back(x);
		}
		deque<int>::iterator it;
		int ans = 0;
		for (int i = 1; i <= n; i++)
		{
			for (int j =1; j <= m; j++)
			{
				scanf("%d", &x);
				for (it = v.begin(); it != v.end(); it++)
				{
					ans++;
					if (*it == x)
					{
						v.erase(it);
						break;
					}
				}
				v.push_front(x);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}