Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input
The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.
It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).
Output
Output a single integer — the minimum number of plants to be replanted.
Examples
input
3 2 2 1 1 2.0 1 3.100
output
1
input
3 3 1 5.0 2 5.5 3 6.0
output
0
input
6 3 1 14.284235 2 17.921382 1 20.328172 3 20.842331 1 25.790145 1 27.204125
output
2
Note
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed.
这个题跟n，跟后边那一列没有一点关系，只考虑前一列，那么这就是个LIS，直接摸板过。
#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define debug(n) printf("%d_________________________________
",n);
#define speed ios_base::sync_with_stdio(0)
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) (a>b?a:b)
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
#define pb(n) push_back(n)
//--------------------------------constant----------------------------------//
#define INF 0x3f3f3f3f
#define maxn 10000000
#define esp 1e-9
using namespace std;
typedef pair<int,int>PII;
typedef pair<string,int>PSI;
typedef long long ll;
//___________________________Dividing Line__________________________________//
#include <iostream>
#include<cstring>
using namespace std;
const int nmax=1000000+5;
int a[nmax];
int dp[nmax];
int main() {
int n;
double m;
while(cin>>n>>m){
for(int i=0;i<n;i++){
cin>>a[i]>>m;
dp[i]=1;
}
int ans=1;
for(int i=1;i<n;i++){
for(int j=0;j<i;j++){
if(a[j]<=a[i]){
dp[i]=max(dp[j]+1,dp[i]);
}
ans=max(ans,dp[i]);
}
}
cout<<n-ans<<endl;
}
return 0;
}