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

public class Main {
	static int mod = 1000000007;
 	public static void main(String[] args){
		FastScanner scanner = new FastScanner();
		int n = scanner.nextInt();
		int[] a = new int[n];
		int[] b = new int[n];
		for(int i = 0; i < n; i++){
			a[i] = scanner.nextInt();
			b[i] = a[i];
		}
		Arrays.sort(a);
		int ans = 0;
		ArrayList<Integer> list = new ArrayList<Integer>();
		for(int i = n-1; i >= 0; i--){
			int pos = i;
			for(int j = 0; j < n; j++){
				if(b[j] != 0 && a[pos] == b[j]){
					list.add(j+1);
					b[j] = 0;
				}
			}
			ans += (n-1-i)*a[pos]+1;
		}
		System.out.println(ans);
		for(int i = 0; i < list.size(); i++){
			System.out.print(list.get(i)+" ");
		}
	}
	static class UnionFind {
		int[] parent;
		public UnionFind(int size) {
			parent = new int[size];
			Arrays.fill(parent, -1);
		}
		public boolean unite(int x, int y) {
			x = root(x);
			y = root(y);
			if (x != y) {
				if (parent[y] < parent[x]) {
					int tmp = y;
					y = x;
					x = tmp;
				}
				parent[x] += parent[y];
				parent[y] = x;
				return true;
			}
			return false;
		}
		public boolean same(int x, int y) {
			return root(x) == root(y);
		}
		public int root(int x) {
			return parent[x] < 0 ? x : (parent[x] = root(parent[x]));
		}
		public int size(int x) {
			return -parent[root(x)];
		}
	}
	static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        //return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        //return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
		    //return first * 1.0 / second > p.first * 1.0 / p.second ? 1 : -1; // first/secondの昇順にソート
		    //return first * 1.0 / second < p.first * 1.0 / p.second ? 1 : -1; // first/secondの降順にソート
				//return second * 1.0 / first > p.second * 1.0 / p.first ? 1 : -1; // second/firstの昇順にソート
		    //return second * 1.0 / first < p.second * 1.0 / p.first ? 1 : -1; // second/firstの降順にソート
				//return Math.atan2(second, first) > Math.atan2(p.second, p.first) ? 1 : -1; // second/firstの昇順にソート
				//return first + second > p.first + p.second ? 1 : -1; //first+secondの昇順にソート
				return first + second < p.first + p.second ? 1 : -1; //first+secondの降順にソート
				//return first - second < p.first - p.second ? 1 : -1; //first+secondの降順にソート
		}
  }

	private static class FastScanner {
		private final InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
				if (ptr < buflen) {
						return true;
				}else{
						ptr = 0;
						try {
								buflen = in.read(buffer);
						} catch (IOException e) {
								e.printStackTrace();
						}
						if (buflen <= 0) {
								return false;
						}
				}
				return true;
		}
		private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
		private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
		public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
		public String next() {
				if (!hasNext()) throw new NoSuchElementException();
				StringBuilder sb = new StringBuilder();
				int b = readByte();
				while(isPrintableChar(b)) {
						sb.appendCodePoint(b);
						b = readByte();
				}
				return sb.toString();
		}
		public long nextLong() {
				if (!hasNext()) throw new NoSuchElementException();
				long n = 0;
				boolean minus = false;
				int b = readByte();
				if (b == '-') {
						minus = true;
						b = readByte();
				}
				if (b < '0' || '9' < b) {
						throw new NumberFormatException();
				}
				while(true){
						if ('0' <= b && b <= '9') {
								n *= 10;
								n += b - '0';
						}else if(b == -1 || !isPrintableChar(b)){
								return minus ? -n : n;
						}else{
								throw new NumberFormatException();
						}
						b = readByte();
				}
		}
		public int nextInt() {
				long nl = nextLong();
				if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
				return (int) nl;
		}
		public double nextDouble() { return Double.parseDouble(next());}
	}
}