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: Runtime Error)

#include<bits/stdc++.h> 
#define INF 0x3f3f3f  
using namespace std;
int dp[20010];//dp[i]表示长度为i+1的子序列末尾元素最小值；   
int a[20010];  
int main()  
{  
    int n;  
    while(scanf("%d",&n)!=EOF)  
    {  
        for(int i=0;i<n;i++)  
        {  
            scanf("%d",&a[i]);  
            dp[i]=INF;//不可以用memset对数组赋值INF,只能赋值0或-1；  
                      //可以用fill(dp,dp+n,INF);   
        }  
        for(int i=0;i<n;i++)  
        {  
            *lower_bound(dp,dp+n,a[i])=a[i];//找到>=a[i]的第一个元素，并用a[i]替换；   
        }  
        printf("%d\n",n-(lower_bound(dp,dp+n,INF)-dp)+1);//找到第一个INF的地址减去首地址就是最大子序列的长度；   
    }  
    return 0;  
}