Problem: Recently Vasya decided to improve his pistol shooting skills. Today his coach offered him the following exercise. He placed n cans in a row on a table. Cans are numbered from left to right from 1 to n. Vasya has to knock down each can exactly once to finish the exercise. He is allowed to choose the order in which he will knock the cans down.

Vasya knows that the durability of the i-th can is a_i. It means that if Vasya has already knocked x cans down and is now about to start shooting the i-th one, he will need (a_i ⋅ x + 1) shots to knock it down. You can assume that if Vasya starts shooting the i-th can, he will be shooting it until he knocks it down.

Your task is to choose such an order of shooting so that the number of shots required to knock each of the n given cans down exactly once is minimum possible.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 1 000) — the number of cans.

The second line of the input contains the sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1 000), where a_i is the durability of the i-th can.

Output

In the first line print the minimum number of shots required to knock each of the n given cans down exactly once.

In the second line print the sequence consisting of n distinct integers from 1 to n — the order of indices of cans that minimizes the number of shots required. If there are several answers, you can print any of them.

Examples

Input


3
20 10 20


Output


43
1 3 2 


Input


4
10 10 10 10


Output


64
2 1 4 3 


Input


6
5 4 5 4 4 5


Output


69
6 1 3 5 2 4 


Input


2
1 4


Output


3
2 1 

Note

In the first example Vasya can start shooting from the first can. He knocks it down with the first shot because he haven't knocked any other cans down before. After that he has to shoot the third can. To knock it down he shoots 20 ⋅ 1 + 1 = 21 times. After that only second can remains. To knock it down Vasya shoots 10 ⋅ 2 + 1 = 21 times. So the total number of shots is 1 + 21 + 21 = 43.

In the second example the order of shooting does not matter because all cans have the same durability.

Solution (in JAVA):

Solution:
/*      MOHD SADIQ        */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;

public class Main {

    private static  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    private  long longVal() throws IOException{
        return Long.parseLong(br.readLine().trim());
    }
    private  int intVal() throws IOException{
        return Integer.parseInt(br.readLine().trim());
    }
    private  double doubleVal() throws IOException{
        return Double.parseDouble(br.readLine().trim());
    }
    private  int gcdExtended(int a, int b, int x, int y)
    {
        if (a == 0)
        {
            x = 0;
            y = 1;
            return b;
        }

        int x1=1, y1=1;
        int gcd = gcdExtended(b%a, a, x1, y1);

        x = y1 - (b/a) * x1;
        y = x1;

        return gcd;
    }
    private  long pow(long x,long y){
        long res =1;
        while (y>0){
            if(y%2!=0){
                res = res *x;
            }
            y/=2;
            x=x*x;
        }
        return res;
    }

    private long powMod(long x, long y, long mod) {
        long res = 1;
        while (y > 0) {
            if (y % 2 != 0) {
                res = (res * x) % mod;
            }
            y /= 2;
            x = (x * x) % mod;
        }
        return res;
    }

    private boolean[] sieve(int n){
        boolean[] primes = new boolean[n + 1];
        int i;
        Arrays.fill(primes,true);
        for(i = 2; i * i <= n; ++i) {
            if (primes[i]) {
                for(int p = i * i; p <= n; p += i) {
                    primes[p] = false;
                }
            }
        }
        return primes;
    }

    private int gcd(int a,int b){
        return gcdExtended(a,b,1,1);
    }

    private  int[] intArr() throws IOException{
        String[] s = br.readLine().trim().split(" ");
        int n = s.length;
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(s[i]);
        }
        return a;
    }
    private  long[] longArr() throws IOException {
        String[] s = br.readLine().trim().split(" ");
        int n = s.length;
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = Long.parseLong(s[i]);
        }
        return a;
    }
    private  double[] doubleArr() throws IOException {
        String[] s = br.readLine().trim().split(" ");
        int n = s.length;
        double[] a = new double[n];
        for (int i = 0; i < n; i++) {
            a[i] = Double.parseDouble(s[i]);
        }
        return a;
    }
    private String[] stringArr() throws IOException{
        String[] s = br.readLine().trim().split(" ");
        return s;
    }
    private  char[] charArr() throws IOException{
        return br.readLine().trim().toCharArray();
    }

    private static long mod = (long)Math.pow(10,9)+7;


    private static String solve() throws IOException {

        Main obj = new Main();
        int n = obj.intVal();
        int[] arr = obj.intArr();
//        int[] a = new int[1001];
        ArrayList<Integer>[] al = new ArrayList[1001];
        for(int i=0;i<=1000;i++)
            al[i] = new ArrayList<>();
        for(int i=0;i<n;i++){
            al[arr[i]].add(i+1);
        }
        int c = 0;
        long sum = 0;
        StringBuffer ans = new StringBuffer();
        for(int i=1000;i>0;i--){
            if(al[i].size()!=0){
                for(int j=0;j<al[i].size();j++){
                    sum += (c*i)+1;
                    ans.append(al[i].get(j)+" ");
                    c++;
                }
            }
        }
        return sum+"\n"+ans;
    }


    public static void main(String[] args) throws IOException {
        Main ob = new Main();
//        int total_test_case = ob.intVal();
        int total_test_case = 1;
        StringBuffer ans = new StringBuffer();
        for(int test_case =0;test_case<total_test_case;test_case++){
            ans.append(solve()+"\n");
//            System.out.println(solve(a));
        }
        System.out.print(ans);
    }
}