Codeforces Round #335 (Div. 2)-C Sorting Railway Cars（最长上升子序列）

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

54 1 2 5 3

output

2

input

44 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：给你一个无序系列，每次可以将一个数往两端放，问你放的最少次数使得该序列有序。

题解：直接求一发最长上升子序列，然后总长度-该长度即可。。。

#include<set>      #include<map>         #include<stack>                #include<queue>                #include<vector>        #include<string>     #include<time.h>    #include<math.h>                #include<stdio.h>                #include<iostream>                #include<string.h>                #include<stdlib.h>        #include<algorithm>       #include<functional>        using namespace std;                #define ll long long          #define inf 1000000000           #define mod 1000000007                #define maxn  205000    #define lowbit(x) (x&-x)                #define eps 1e-9  int a[maxn],b[maxn];int main(void){int n,i,ans;scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);for(i=1;i<=n;i++)b[a[i]]=b[a[i]-1]+1;for(i=1;i<=n;i++)ans=max(ans,b[i]);printf("%d\n",n-ans);return 0;}