Codeforces Round #335 (Div. 2)C. Sorting Railway Cars

C. Sorting Railway Cars

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

An infinitely long railway has a train consisting of n cars, numbered from 1 to n (the
numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning
of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?

Input

The first line of the input contains integer n (1 ≤ n ≤ 100 000) —
the number of cars in the train.

The second line contains n integers pi (1 ≤ pi ≤ n, pi ≠ pj if i ≠ j) —
the sequence of the numbers of the cars in the train.

Output

Print a single integer — the minimum number of actions needed to sort the railway cars.

Examples

input

5
4 1 2 5 3

output

2

input

4
4 1 3 2

output

2

Note

In the first sample you need first to teleport the 4-th car, and then the 5-th
car to the end of the train.

题意：给出乱序的1~n个数每次可以寻去一个数将其放到最后面或者放到最前面问最少需要多少次可以将这个序列变得有序。

思路：求出所给的序列中最长的连续的数对剩下的数进行两种操作即可。最终答案为总数-最长的连续的数的个数

/* ***********************************************
Author : ryc
Created Time : 2016-08-22 Monday
File Name : E:\acm\codeforces\335C.cpp
Language : c++
Copyright 2016 ryc All Rights Reserved
************************************************ */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<list>
#include<vector>
#include<map>
#include<stack>
#include<string>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=1000010;
int num[maxn],sum[maxn];
int main()
{
int n;cin>>n;
for(int i=1;i<=n;++i){
scanf("%d",&num[i]);
sum[num[i]]=sum[num[i]-1]+1;
}
int ans=0;
for(int i=1;i<=n;++i){
ans=max(ans,sum[i]);
}
printf("%d\n",n-ans);
return 0;
}