JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines.

Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource.

With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.

Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.

The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones.

But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.

For example, the roads in Figure I are forbidden.



In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^

Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.

For each test case, output the result in the form of sample.
You should tell JGShining what's the maximal number of road(s) can be built.

JGShining（極光炫影）

/*此題資料量略大，所以用dp直接求果斷TLE了，
所以經這道題我學習了一下優化的演算法，
也就是在計算以當前項結尾的最長遞增序列時使用二分查詢，
一次降低時間複雜度*/
//注意：輸出時單數road複數要用roads，此細節wa了無數次。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <deque>
using namespace std;
int r[500001];
//b[i]儲存的是長度為i的遞增序列的最小末尾項
int b[500001];
deque<int> q;

int DP(int n)
{
    int i,left=1,mid=0,right=1,len=1;
    b[1]=r[1];
    for(i=2;i<=n;i++)//計算以每一項結尾的最長遞增序列
    {
        right=len;
        left=1;
        while(left<=right)//二分查詢
        {
            //對b陣列，從下標的中值也就是長度開始查詢，
            //如果當前項比b[mid]大，
            //那說明當前項可以插到長度比maxlen/2大的序列的尾部，
            //於是將範圍縮小到後半段，接著使用二分，
            //反之亦然
            mid=(left+right)/2;
            if(b[mid]<r[i])
                left=mid+1;
            else
                right=mid-1;
        }
        b[left]=r[i];
        if(left>len)
            len+=1;
    }
    return len;
}

int main()
{
    int n,t=0;
    while(cin>>n)
    {
        t++;
        memset(b,0,sizeof(dp));
        int i,ans;
        for(i=0;i<n;i++)
        {
            int rr,pp;
            scanf("%d%d",&rr,&pp);
            r[rr]=pp;
        }
        ans=DP(n);
        cout<<"Case "<<t<<":"<<endl;
        if(ans==1)
        cout<<"My king, at most "<<ans<<" road can be built."<<endl;
        else
        cout<<"My king, at most "<<ans<<" roads can be built."<<endl;
        cout<<endl;
    }
    return 0;
}