This problem is same as the previous one, but has larger constraints.

Shiro's just moved to the new house. She wants to invite all friends of her to the house so they can play monopoly. However, her house is too small, so she can only invite one friend at a time.

For each of the n days since the day Shiro moved to the new house, there will be exactly one cat coming to the Shiro's house. The cat coming in the i-th day has a ribbon with color u_i. Shiro wants to know the largest number x, such that if we consider the streak of the first x days, it is possible to remove exactly one day from this streak so that every ribbon color that has appeared among the remaining x - 1 will have the same number of occurrences.

For example, consider the following sequence of u_i: [2, 2, 1, 1, 5, 4, 4, 5]. Then x = 7 makes a streak, since if we remove the leftmost u_i = 5, each ribbon color will appear exactly twice in the prefix of x - 1 days. Note that x = 8 doesn't form a streak, since you must remove exactly one day. 

Since Shiro is just a cat, she is not very good at counting and needs your help finding the longest streak.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the total number of days.

The second line contains n integers u_1, u_2, …, u_n (1 ≤ u_i ≤ 10^5) — the colors of the ribbons the cats wear. 

Output

Print a single integer x — the largest possible streak of days.

Examples

Input


13
1 1 1 2 2 2 3 3 3 4 4 4 5


Output


13

Input


5
10 100 20 200 1


Output


5

Input


1
100000


Output


1

Input


7
3 2 1 1 4 5 1


Output


6

Input


6
1 1 1 2 2 2


Output


5

Note

In the first example, we can choose the longest streak of 13 days, since upon removing the last day out of the streak, all of the remaining colors 1, 2, 3, and 4 will have the same number of occurrences of 3. Note that the streak can also be 10 days (by removing the 10-th day from this streak) but we are interested in the longest streak.

In the fourth example, if we take the streak of the first 6 days, we can remove the third day from this streak then all of the remaining colors 1, 2, 3, 4 and 5 will occur exactly once.
"""
    Satwik_Tiwari ;) .
    24th Sept , 2020  - Thursday
"""

#===============================================================================================
#importing some useful libraries.


from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


# from itertools import *
# from heapq import *
from math import gcd, factorial,floor,ceil

from copy import deepcopy
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
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
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")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

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

mod = 10**9+7
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 pp in range(t):
        solve(pp)
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
    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 solve(case):

    n = int(inp())
    a = lis()

    cnt = [0] * (10**5+1)
    ll = 0
    ans = 0
    have = set()

    revcnt = {0:n}
    for i in range(1,10**5+1):
        revcnt[i] = 0
    for i in range(n):
        if(cnt[a[i]] == 0):
            ll+=1
        else:
            if(revcnt[cnt[a[i]]] == 1):
                have.remove(cnt[a[i]])
        revcnt[cnt[a[i]]]-=1
        cnt[a[i]] +=1
        revcnt[cnt[a[i]]]+=1
        have.add(cnt[a[i]])
        # print(revcnt)
        if(len(have)==2):
            new = list(have)
            new = sorted(new)
            # print(new)
            if(revcnt[new[0]]>1 and revcnt[new[1]]>1):
                continue
            if(len(have) == 2):
                if(((new[1]-new[0]) == 1 and revcnt[new[1]]==1) or (new[0]==1 and revcnt[1] ==1)):
                    ans = i+1
        elif(len(have) == 1 and (revcnt[1] == ll or ll==1)):
            ans = i+1
    print(ans)



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









