# Into Blocks (easy version)

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

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, tt;
cin >> n >> tt;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
const int MAX = 200010;
vector<int> first(MAX, -1), last(MAX, -1), cnt(MAX, 0);
for (int i = 0; i < n; i++) {
if (first[a[i]] == -1) {
first[a[i]] = i;
}
last[a[i]] = i;
++cnt[a[i]];
}
vector<pair<int, int>> e;
for (int i = 0; i < MAX; i++) {
if (first[i] != -1) {
e.emplace_back(first[i], i);
}
}
sort(e.begin(), e.end());
int ans = 0;
int ls = -1;
int cmax = 0;
for (int i = 0; i < (int) e.size(); i++) {
int id = e[i].second;
if (first[id] > ls) {
ans += cmax;
cmax = cnt[id];
ls = last[id];
} else {
cmax = max(cmax, cnt[id]);
ls = max(ls, last[id]);
}
}
ans += cmax;
cout << n - ans << '\n';
return 0;
}