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?

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 p_i (1 ≤ p_i ≤ n, p_i ≠ p_j if i ≠ j) — the sequence of the numbers of the cars in the train.

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

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.

思路：因为最后的答案一定是1，2，3，4，。。。。n；

　　　所以我本来想求最长公共子序列，发现时间复杂度太大；

　　　后来想因为一定上升只要求最长连续上升即可；

#include<bits/stdc++.h>

using namespace std;

#define ll __int64

#define esp 1e-13

const int N=1e5+,M=1e6+,inf=1e9+,mod=;

int a[N];

int dp[N];

int flag[N];

int main()

{

    int x,y,z,i,t;

    scanf("%d",&x);

    for(i=;i<=x;i++)

    scanf("%d",&y),flag[y]=i;

    dp[]=;

    for(i=;i<=x;i++)

    {

        dp[i]=;

        if(flag[i]>flag[i-])

        dp[i]=dp[i-]+;

    }

    int ans=;

    for(i=;i<=x;i++)

        ans=max(ans,dp[i]);

    printf("%d\n",x-ans);

    return ;

}