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:
import sys
import os
from io import IOBase, BytesIO
# import heapq
import math
import collections
# import itertools
# import bisect
mod = 10 ** 9 + 7
pie = 3.1415926536
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**6)
# fact=[1]
# for i in range(1,1000001):
#    fact.append((fact[-1]*i)%mod)
# ifact=[0]*1000001
# ifact[1000000]=pow(fact[1000000],mod-2,mod)
# for i in range(1000000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
# from random import randint as rn
# from Queue import Queue as Q


def modinv(n, p):
    return pow(n, p-2, p)


def ncr(n, r, p):  # for using this uncomment the lines calculating fact and ifact
    t = ((fact[n])*((ifact[r]*ifact[n-r]) % p)) % p
    return t


def ain():  # takes array as input
    return list(map(int, sin().split()))


def sin():
    return input().strip()


def GCD(x, y):
    while(y):
        x, y = y, x % y
    return x


def read2DIntArray(row, col):
    arr = []
    for i in range(0, row):
        temp = list(map(int, sin().split()))
        arr.append(temp)

    return arr


def read2DCharArray(row, col):
    arr = []
    for i in range(0, row):
        temp = str(sin())
        arr.append(temp)

    return arr


"""****************** SMALLEST NO. BY REARRANGING DIGITS OF n (WITHOUT TRAILING ZEROS) *********************"""


def smallestNumber(n):
    lst = list(str(n))
    lst.sort()

    tmp = ""
    for i, n in enumerate(lst):
        if (n != '0'):
            tmp = lst.pop(i)
            break

    return str(tmp) + ''.join(lst)


"""*********************** GENERATE ALL PRIME NUMBERS SMALLER THAN OR EQUAL TO n ***************************"""


def SieveOfEratosthenes(n):

    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):

        if (prime[p] == True):
            for i in range(p * 2, n + 1, p):
                prime[i] = False

        p += 1
    prime[0] = False
    prime[1] = False

    lst = []
    for p in range(n + 1):
        if prime[p]:
            lst.append(p)

    return lst


"""*************************************** FIND nCr ********************************************************"""


def nCr(n, r):
    a = 1
    b = 1
    c = 1
    for i in range(1, n + 1):
        c *= i
    for i in range(1, r + 1):
        b *= i
    for i in range(1, n - r + 1):
        a *= i

    return (c // (a * b))


"""************************* GET PRIME FACTORS AND THEIR POWERS FOR AN INTEGER *****************************"""


def sieveOfEratosthenes1(N, s):

    prime = [False] * (N+1)

    for i in range(2, N+1, 2):
        s[i] = 2

    for i in range(3, N+1, 2):
        if (prime[i] == False):

            s[i] = i

            for j in range(i, int(N / i) + 1, 2):
                if (prime[i*j] == False):
                    prime[i*j] = True

                    s[i * j] = i


def generatePrimeFactors(N):

    s = [0] * (N+1)

    sieveOfEratosthenes1(N, s)

    # print("Factor Power")

    curr = s[N]

    cnt = 1

    factors = []
    power = []
    while (N > 1):
        N //= s[N]

        if (curr == s[N]):
            cnt += 1
            continue

        # curr is factor and cnt in the power of this factor
        factors.append(curr)
        power.append(cnt)

        curr = s[N]
        cnt = 1

    return factors, power


"""----------------------------------------------MAIN------------------------------------------------------"""


def main():
    n = int(sin())
    a = ain()

    lst = []
    for i in range(n):
        lst.append((a[i], i + 1))

    lst = sorted(lst, key=lambda x: x[0], reverse=True)

    ans = 0
    for i in range(n):
        idx = lst[i][1]-1
        ans += a[idx] * i + 1

    print(ans)
    for i in range(n):
        print(lst[i][1], end=" ")


"""--------------------------------------------------------------------------------------------------------"""

# Python 2 and 3 footer by Pajenegod and j1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange


BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2),
                   super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill():
            pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill()
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def input(): return sys.stdin.readline().rstrip('\r\n')


if __name__ == '__main__':
    main()
# threading.Thread(target=main).start()