This is an easier version of the next problem. In this version, $q = 0$.

A sequence of integers is called nice if its elements are arranged in blocks like in $[3, 3, 3, 4, 1, 1]$. Formally, if two elements are equal, everything in between must also be equal.

Let's define difficulty of a sequence as a minimum possible number of elements to change to get a nice sequence. However, if you change at least one element of value $x$ to value $y$, you must also change all other elements of value $x$ into $y$ as well. For example, for $[3, 3, 1, 3, 2, 1, 2]$ it isn't allowed to change first $1$ to $3$ and second $1$ to $2$. You need to leave $1$'s untouched or change them to the same value.

You are given a sequence of integers $a_1, a_2, \ldots, a_n$ and $q$ updates.

Each update is of form "$i$ $x$" — change $a_i$ to $x$. Updates are not independent (the change stays for the future).

Print the difficulty of the initial sequence and of the sequence after every update.


-----Input-----

The first line contains integers $n$ and $q$ ($1 \le n \le 200\,000$, $q = 0$), the length of the sequence and the number of the updates.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 200\,000$), the initial sequence.

Each of the following $q$ lines contains integers $i_t$ and $x_t$ ($1 \le i_t \le n$, $1 \le x_t \le 200\,000$), the position and the new value for this position.


-----Output-----

Print $q+1$ integers, the answer for the initial sequence and the answer after every update.


-----Examples-----
Input
5 0
3 7 3 7 3

Output
2

Input
10 0
1 2 1 2 3 1 1 1 50 1

Output
4

Input
6 0
6 6 3 3 4 4

Output
0

Input
7 0
3 3 1 3 2 1 2

Output
4

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import itertools\nimport sys\n\ndef __starting_point():\n    n, q = map(int, input().split())\n    blocks = list(map(int, input().split()))\n    # create empty dictionary - unordered collection\n    dif = dict()\n\n    for i in range(n):\n        if blocks[i] in dif:\n            dif[blocks[i]][1] = i\n            dif[blocks[i]][2] = dif[blocks[i]][2] + 1\n        elif blocks[i] not in dif:\n            dif[blocks[i]] = [i, i, 1]\n\n    rez = 0\n    end = -1\n    maxi = -1\n    for i in range(n):\n        if dif[blocks[i]][1] >= end:\n            end = dif[blocks[i]][1]\n        if dif[blocks[i]][2] >= maxi:\n            maxi = dif[blocks[i]][2]\n        if i == end:\n            rez  = rez + maxi\n            maxi = 0\n    rez = (-1) * rez\n    for i in dif:\n        rez = rez + dif[i][2]\n    print(rez)\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13961