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 PYTHON3):

Solution:
"""
    Satwik_Tiwari ;) .
    28 june , 2020  - Sunday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def bs(a,l,h,x):
    while(l<h):
        # print(l,h)
        mid = (l+h)//2
        if(a[mid] == x):
            return mid
        if(a[mid] < x):
            l = mid+1
        else:
            h = mid
    return l

def sieve(a): #O(n loglogn) nearly linear
    #all odd mark 1
    for i in range(3,((10**6)+1),2):
        a[i] = 1
    #marking multiples of i form i*i 0. they are nt prime
    for i in range(3,((10**6)+1),2):
        for j in range(i*i,((10**6)+1),i):
            a[j] = 0
    a[2] = 1 #special left case
    return (a)


def bfs(g,st):
    visited = [0]*(len(g))
    visited[st] = 1
    queue = []
    queue.append(st)
    new = []
    while(len(queue) != 0):
        s = queue.pop()
        new.append(s)
        for i in g[s]:
            if(visited[i] == 0):
                visited[i] = 1
                queue.append(i)
    return new


def solve():
    n = int(inp())
    a = lis()
    h = []
    for i in range(0,n):
        heappush(h,[-a[i],i+1])
    ans = 0
    cnt = 0
    order = []
    for i in range(0,n):
        elem = heappop(h)
        ans += (-elem[0])*cnt + 1
        cnt+=1
        order.append(elem[1])

    print(ans)
    print(*order)




testcase(1)
# testcase(int(inp()))