Problem:
Score: 100 points

Problem Statement

M-kun is a competitor in AtCoder, whose highest rating is X.

In this site, a competitor is given a kyu (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given:

From 400 through 599: 8-kyu

From 600 through 799: 7-kyu

From 800 through 999: 6-kyu

From 1000 through 1199: 5-kyu

From 1200 through 1399: 4-kyu

From 1400 through 1599: 3-kyu

From 1600 through 1799: 2-kyu

From 1800 through 1999: 1-kyu

What kyu does M-kun have?

Constraints

400 \leq X \leq 1999

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the kyu M-kun has, as an integer.
For example, if he has 8-kyu, print 8.

Sample Input 1

725

Sample Output 1

7

M-kun's highest rating is 725, which corresponds to 7-kyu.

Thus, 7 is the correct output.

Sample Input 2

1600

Sample Output 2

2

M-kun's highest rating is 1600, which corresponds to 2-kyu.

Thus, 2 is the correct output.

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include<complex>
#include<iomanip>
#include<unordered_map>
#define deg(x) cout<<#x<<"="<<x<<endl;
#define fd(x) for(int i=0;i<x;i++)
#define fdd(a,x) for(int i=a;i<x;i++)
#define fx(x,n) for(int i=n-1;i>=x;i--)
#define mst(x) memset(x,0,sizeof(x));
#define fi first
#define se second
#define TB ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ok "ok"
#define HH puts("");
#define ls rt<<1
#define rs rt<<1|1
#define pb(a,b) a.push_back(b);
#define FIN  freopen("1.in","r",stdin);
#define FOUT freopen("1.out","w",stdout);
using namespace std;
typedef long long ll;typedef long long LL;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<string,int> pss;
typedef unsigned long long ull;
const ll mod=1e9+7;
const int inf=1e9+7;
const double eps=1e-6;
ll qpow(ll a,ll b,ll m){ ll r=1;a%=m;for(;b;b>>=1){if(b&1)r=r*a%m;a=a*a%m;}return (r+m)%m;}
const double pi=acos(-1);


ll a[100];
ll v[100];

int main()
{
    TB
    int n;
    cin>>n;
    ll qian=1000,gu=0;
    fd(n) cin>>a[i];
    int i=0,flag=0;
    while(i<n){
        int j=i;
        while(j+1<n&&a[j]>=a[j+1]) j++;
        v[j]=2;
        i=j;j=i;
        if(j==n-1) break;
        while(j+1<n&&a[j]<=a[j+1]) j++;
        v[j]=1;
        flag++;
        i=j;i++;
    }
    //fd(n) deg(v[i])
    for(int i=0;i<n;i++){
        if(v[i]==2&&flag){
            flag--;
            gu+=qian/a[i];
            qian%=a[i];
        }else if(v[i]==1){
            qian+=gu*a[i];
            gu=0;
        }
    }
    cout<<qian<<'\n';
    return 0;
}