题干：

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.

题目大意：

   给出n棵树，每颗树都属于某种品种，品种按照1-m标号。每个蔬菜可视为一个点，现在给出了每棵树的品种和位置（实数）（给出的顺序按照x坐标升序排序），问如何用最小的移动次数（挖出来种下去）使得这n棵树的品种按照递增的序号牌号。（使得菜的种类能按从左到右非递减排序。）

解题报告：

   考虑该问题的对偶问题，也就是最大的不挪动次数，也就是最大非递减树数。于是跑一个nlogn的LIS就可以了（貌似n^2的也能过、、）

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int n,m;
int a[MAX];
double tmp;
int b[MAX];
int DP() {
	b[1] = a[1];
	int len = 1;
	for(int i = 2; i<=n; i++) {
		if(a[i] >= b[len]) b[++len] = a[i];
		else {
			int pos = upper_bound(b+1,b+len+1,a[i]) - b;
			b[pos] = a[i];
		}
	}
	return len;
}
int main()
{
	cin>>n>>m;
	for(int i = 1; i<=n; i++) scanf("%d%lf",a+i,&tmp);
	printf("%d\n",n-DP());

	return 0 ;
 }