Problem:
Score : 500 points

Problem Statement

You are given a sequence (P_1,P_2,...,P_N) which is a permutation of the integers from 1 through N.
You would like to sort this sequence in ascending order by repeating the following operation:

Choose an element in the sequence and move it to the beginning or the end of the sequence.

Find the minimum number of operations required.
It can be proved that it is actually possible to sort the sequence using this operation.

Constraints

1 \leq N \leq 2\times 10^5

(P_1,P_2,...,P_N) is a permutation of (1,2,...,N).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1
:
P_N

Output

Print the minimum number of operations required.

Sample Input 1

4
1
3
2
4

Sample Output 1

2

For example, the sequence can be sorted in ascending order as follows:

Move 2 to the beginning. The sequence is now (2,1,3,4).

Move 1 to the beginning. The sequence is now (1,2,3,4).

Sample Input 2

6
3
2
5
1
4
6

Sample Output 2

4

Sample Input 3

8
6
3
1
2
7
4
8
5

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <array>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <unordered_map>
#include <unordered_set>
*/

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())

typedef long long ll;
const ll inf = 2e9 + 7;
const int MAX_N = 5e5 + 5;
int dx[] = { -1,0,1,0 }, dy[] = { 0,1,0,-1 };
vector<ll> prime;

ll inv[MAX_N], fac[MAX_N];

template <class T = int> T in() { T x; cin >> x; return (x); }
inline ll GCD(ll a, ll b) { ll c; while (b != 0) { c = a % b; a = b; b = c; }return a; }
inline ll LCM(ll a, ll b) { return a * b / GCD(a, b); }
inline ll POW(ll a, ll b) { ll c = 1; while (b > 0) { if (b & 1) { c = a * c%inf; }a = a * a%inf; b >>= 1; }return c; }
inline void _nCr() { fac[0] = 1; for (int i = 1; i < MAX_N; i++) { fac[i] = fac[i - 1] * i%inf; }for (int i = 0; i < MAX_N; i++) { inv[i] = POW(fac[i], inf - 2); } }
inline ll nCr(ll n, ll r) { return (fac[n] * inv[r] % inf)*inv[n - r] % inf; }
inline void PRI(ll n) { bool a[n + 1]; for (int i = 0; i < n + 1; i++) { a[i] = 1; }for (int i = 2; i < n + 1; i++) { if (a[i]) { prime.pb(i); ll b = i; while (b <= n) { a[b] = 0; b += i; } } } }

typedef pair<int, pair<int, int>> edge;

class UnionFind {
private:
	vector<int> par;
public:
	UnionFind(int N) { par = vector<int>(N, -1); }
	int find(int x);
	ll size(int x);
	void unite(int x, int y);
	bool same(int x, int y);
};



class Kruskal {
private:
	UnionFind *uf;
	vector<edge> e;
public:
	vector<edge> mst;
	Kruskal(int N) { uf = new UnionFind(N); }
	void add(int x, int y, ll z);
	void run();
};



//----UnionFind-------------------------------
int UnionFind::find(int x) {
	if (par[x] < 0) return x;
	else return par[x] = find(par[x]);
}

ll UnionFind::size(int x) {
	return -par[find(x)];
}

void UnionFind::unite(int x, int y) {
	x = find(x);
	y = find(y);

	//大きい方に小さい方をくっ付ける
	if (size(x) < size(y)) swap(x, y);
	par[x] += par[y];
	par[y] = x;
}

bool UnionFind::same(int x, int y) {
	x = find(x);
	y = find(y);
	return x == y;
}



//----Kruskal-------------------------------
void Kruskal::add(int x, int y, ll z) {
	//x < y
	if (x > y) swap(x, y);
	e.push_back({ z,{x,y} });
}

void Kruskal::run() {
	sort(e.begin(), e.end());
	e.erase(unique(e.begin(), e.end()), e.end());
	for (auto x : e) {
		if (uf->same(x.second.first, x.second.second)) {
			continue;
		}
		else {
			mst.push_back(x);
			uf->unite(x.second.first, x.second.second);
		}
	}
}

signed main() {
	int n;cin >> n;
	int con[n+1];
	REP (i,n+1) {con[i] = 0;}
	int ma = 1;
	REP (i,n) {
		int a;
		cin >> a;
		con[a] = con[a - 1] + 1;
		ma = max(con[a],ma);
	}
	cout << n - ma << endl;
}