Sasha and Dima want to buy two n-tier cakes. Each cake should consist of n different tiers: from the size of 1 to the size of n. Tiers should go in order from the smallest to the biggest (from top to bottom).

They live on the same street, there are 2 ⋅ n houses in a row from left to right. Each house has a pastry shop where you can buy a cake tier. Unfortunately, in each pastry shop you can buy only one tier of only one specific size: in the i-th house you can buy a tier of the size a_i (1 ≤ a_i ≤ n).

Since the guys carry already purchased tiers, and it is impossible to insert a new tier in the middle of the cake, they agreed to buy tiers from the smallest to the biggest. That is, each of them buys tiers in order: 1, then 2, then 3 and so on up to n.

Initially, Sasha and Dima are located near the first (leftmost) house. Output the minimum distance that they will have to walk in total to buy both cakes. The distance between any two neighboring houses is exactly 1.

Input

The first line of the input contains an integer number n — the number of tiers in each cake (1 ≤ n ≤ 10^5).

The second line contains 2 ⋅ n integers a_1, a_2, ..., a_{2n} (1 ≤ a_i ≤ n), where a_i is equal to the size of the tier, which can be bought in the i-th house. Remember that in each house you can buy only one tier. It is guaranteed that every number from 1 to n occurs in a exactly two times.

Output

Print one number — the minimum distance that the guys have to walk in total to buy both cakes. Guys can be near same house at the same time. They begin near the first (leftmost) house. Each of the guys should buy n tiers in ascending order of their sizes.

Examples

Input


3
1 1 2 2 3 3


Output


9


Input


2
2 1 1 2


Output


5


Input


4
4 1 3 2 2 3 1 4


Output


17

Note

In the first example, the possible optimal sequence of actions is:

  * Sasha buys a tier of size 1 near the 1-st house (a_1=1); 
  * Dima goes to the house 2; 
  * Dima buys a tier of size 1 near the 2-nd house (a_2=1); 
  * Sasha goes to the house 4; 
  * Sasha buys a tier of size 2 near the 4-th house (a_4=2); 
  * Sasha goes to the house 5; 
  * Sasha buys a tier of size 3 near the 5-th house (a_5=3); 
  * Dima goes to the house 3; 
  * Dima buys a tier of size 2 near the 3-rd house (a_3=2); 
  * Dima goes to the house 6; 
  * Dima buys a tier of size 3 near the 6-th house (a_6=3). 



So, Sasha goes the distance 3+1=4, and Dima goes the distance 1+1+3=5. In total, they cover a distance of 4+5=9. You can make sure that with any other sequence of actions they will walk no less distance.


import java.io.*;
import java.util.*;

/**
 *  
 * @author akash_2ehuikg
 * Coding is love <3!
 */

public class B {
    public static void main(String args[]) {
        try {
            int n=nextInt();
            TreeMap<Integer,ArrayList<Integer>> tm1=new TreeMap<>();
            for (int i = 0; i < 2*n; i++) {
                int a=nextInt();
                if(tm1.containsKey(a)){
                    ArrayList<Integer> al=tm1.get(a);
                    al.add(i);
                    tm1.put(a,al);
                }else{
                    ArrayList<Integer> al=new ArrayList<>();
                    al.add(i);
                    tm1.put(a,al);
                }
            }
//            System.out.println(""+tm1);
            long ans=0;
            int pos1=0;
            int pos2=0;
            for(int i:tm1.keySet()){
                ArrayList<Integer> al=tm1.get(i);
                int a1=al.get(0);
                int a2=al.get(1);
                if(Math.abs(a1-pos1)+Math.abs(a2-pos2) < Math.abs(a2-pos1)+Math.abs(a1-pos2)){
                    ans+=Math.abs(a1-pos1)+Math.abs(a2-pos2);
                    
                    pos1=a1;
                    pos2=a2;
                }else{
                    ans+= Math.abs(a2-pos1)+Math.abs(a1-pos2);
                    
                    pos1=a2;
                    pos2=a1;
                }
            }
            
            bw.write(ans+"\n");
            bw.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }


    //TEMPLATE
    
    //******************************************************************************
    // Fast I/O 
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st = null;
    static int mod=1000000007;
    static boolean prime[]=new boolean[10000001];

    static String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    
    static int nextInt() {
        return Integer.parseInt(next());
    }
    
    static long nextLong() {
        return Long.parseLong(next());
    }

    static double nextDouble() {
        return Double.parseDouble(next());
    }

    static int b_search(int l,int r,int A[],int key){
        int ans=0;
        l=0;r=1000000000;

        while(l<r){
            //System.out.println(ans);
            ans=(l+r)/2;
            long temp=0;
            for (int i = 0; i < A.length; i++) {
                temp+=(long)Math.ceil((double)A[i]/(double)ans);
            }

            if(temp<=key){
                r=ans;
            }
            else{
                l=ans+1;
                ans=r;
            }
        }

        return ans;
    }

//******************************************************************************
//Sieve
static void sieveOfEratosthenes(boolean prime[]){
        int n=10000000;
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++){ 
            if(prime[p] == true) 
            {
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
    } 

//******************************************************************************
//Modulo and FastExpo
    static long modInverse(long a)
    {
       return power(a, mod - 2,mod);
    }
     
    // To compute x^y under modulo m
    static long power(long x, long y, int p)
    {
        // Initialize result
        long res = 1;     
        // Update x if it is more  
        // than or equal to p
        x = x % p; 
     while (y > 0)
        {
            // If y is odd, multiply x
            // with result
            if((y & 1)==1)
                res = (res * x) % p;
            // y must be even now
            // y = y / 2
            y = y >> 1; 
            x = (x * x) % p; 
        }
        return res;
    }
//******************************************************************************
//GCD
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
         
        return gcd(b%a, a);
    }

    //******************************************************************************
    // Useful user datatype
    
    static class Pair{
        public long x;
        public long y;

        public Pair(long first, long second){
            this.x = first;
            this.y = second;
            }
        @Override
        public String toString() {
        return "(" + x + "," + y + ")";
        }
    }

    static Comparator csort=new Comparator<Pair>(){
                    public int compare(Pair a, Pair b) {
                        if(a.x < b.x)
                        return -1;
                        else if (a.x > b.x)
                        return 1;
                        else if(a.y < b.y)
                        return -1;
                        else if (a.y > b.y)
                        return 1;
                        else
                        return 0;
                    }
                };

    //******************************************************************************
    //N choose K
    public static long choose(long total, long choose){
        if(total < choose)
            return 0;
        if(choose == 0 || choose == total)
            return 1;
        return choose(total-1,choose-1)+choose(total-1,choose);
    }

    //******************************************************************************
    //Permutations
    // simply prints all permutation - to see how it works
    private static void printPermutations( Comparable[] c ) {
            System.out.println( Arrays.toString( c ) );
            while ( ( c = nextPermutation( c ) ) != null ) {
                    System.out.println( Arrays.toString( c ) );
            }
    }

    // modifies c to next permutation or returns null if such permutation does not exist
    private static Comparable[] nextPermutation( final Comparable[] c ) {
            // 1. finds the largest k, that c[k] < c[k+1]
            int first = getFirst( c );
            if ( first == -1 ) return null; // no greater permutation
            // 2. find last index toSwap, that c[k] < c[toSwap]
            int toSwap = c.length - 1;
            while ( c[ first ].compareTo( c[ toSwap ] ) >= 0 )
                    --toSwap;
            // 3. swap elements with indexes first and last
            swap( c, first++, toSwap );
            // 4. reverse sequence from k+1 to n (inclusive) 
            toSwap = c.length - 1;
            while ( first < toSwap )
                    swap( c, first++, toSwap-- );
            return c;
    }

    // finds the largest k, that c[k] < c[k+1]
    // if no such k exists (there is not greater permutation), return -1
    private static int getFirst( final Comparable[] c ) {
            for ( int i = c.length - 2; i >= 0; --i )
                    if ( c[ i ].compareTo( c[ i + 1 ] ) < 0 )
                            return i;
            return -1;
    }

    // swaps two elements (with indexes i and j) in array 
    private static void swap( final Comparable[] c, final int i, final int j ) {
            final Comparable tmp = c[ i ];
            c[ i ] = c[ j ];
            c[ j ] = tmp;
    }
    //******************************************************************************
    // DSU
    //Maintain an Array A with N elements and Array size for size of each set, intialize A[i]=i and size[i]=1

    static int root(int A[],int i){
        while(A[i]!=i){
            A[i]=A[A[i]];
            i=A[i];
        }
        return i;
    }

    static boolean find(int A[],int a,int b){
        if(root(A,a)==root(A,b))return true;
        else return false;
    }

    static void union(int A[],int size[],int a,int b){
        int ra=root(A,a);
        int rb=root(A,b);

        if(ra==rb)return;

        if(size[ra]<size[rb]){
            A[ra]=A[rb];
            size[rb]+=size[ra];
        }else{
            A[rb]=A[ra];
            size[ra]+=size[rb];
        }
    }

    //**************************************************************************
    //BinarySearch
    /* binary search for 5:
                v-- lower bound
         1 2 3 4 5 5 5 6 7 9
                      ^-- upper bound
        
            binary search for 8
                        v-- lower bound
       1 2 3 4 5 5 5 6 7 9
                        ^-- upper bound
    */
    //For finding greater than 
    static int upper_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid;

        while (first < last) {
            mid =first+(last - first)/2; 

            if (A[mid] <= key)  
                first = mid + 1; 
            else 
                last = mid;
        }

        return first;
    }
    
    //For finding greater than equal to
    static int lower_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid=0;
        
        while (first < last) {
            mid = first + ((last - first) >> 1); 
            if (A[mid] < key)  
                first = mid + 1; 
            else 
                last = mid;
        }
        
        return first;
    }
    //**************************************************************************
}
