//sakusen
//dp[M][x] B_Mまで撮影し終え、しかもそのとき機会が一xにあるときの層移動コストの最小値
//B_M=A_xが必要 そうでないならてきとうに998244353(とてもおおきい)にでもしとけ

#include<bits/stdc++.h>
using namespace std;

int main(){
    int N,Q;
    cin >> N >> Q;
    vector<int> A(N+1);
    vector<int> B(Q+1);
    for(int i=1;i<=N;i++){
        cin >> A[i];
    }
    for(int i=1;i<=Q;i++){
        cin >> B[i];
    }
    vector<vector<int>> dp(Q+1, vector<int>(N+1));
    for(int i=1;i<=N;i++){
        if(A[i] == B[1]){
            dp[1][i] = i-1;
        } else {
            dp[1][i] = 998244353;
        }
    }
    int min_temp = 998244353;
    int cost = 0;
    for(int m=2;m<=Q;m++){
        for(int x=1;x<=N;x++){
            //dp[m][x]求めるよ
            if(A[x] != B[m]){
                dp[m][x] = 998244353;
            } else {
                min_temp = 998244353;
                for(int x_0=1;x_0<=N;x_0++){
                    cost = dp[m-1][x_0] + abs(x-x_0);
                    if(cost < min_temp){
                        min_temp = cost;
                    }
                }
                dp[m][x] = min_temp;
            }
        }
    }

    int min = 998244353;
    for(int x=1;x<=N;x++){
        if(dp[Q][x] < min){
            min = dp[Q][x];
        }
    }
    cout << min << endl;
}