A. Sorting Railway Cars

A. 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.

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大意：有n个数字组成的序列，每次可以把序列中的某个数放到开头或者尾部，问最少需要多少次操作才能使序列从小到大排列

分析：我们只需要找到最大的一段不需要移动的序列//即连续的差1的序列  就可以用n-ans得到答案

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100005
using namespace std;

int max(int a,int b)
{
    if(a>b)return a;
    else return b;
}

int main()
{
    int n;
    cin>>n;
    int pos[maxn],a[maxn];
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&a[i]);
        pos[a[i]]=i;
    }
    int cnt=1,ans=1;
    for(int i=2;i<=n;++i)
    { 
        if(pos[i]>pos[i-1])cnt++;
        else cnt=1;
        ans=max(ans,cnt);
    }
    cout<<n-ans;
    puts("");
    return 0;
}