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: Accepted)

//#define _GLIBCXX_DEBUG
#include<bits/stdc++.h>

#define PI 3.14159265359
using namespace std;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define bit(n,k) (((ll)n>>(ll)k)&1) /*nのk bit目*/
#define pb push_back
#define eb emplace_back
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const long long INF= 1e+18+1;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll> >vvl;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> T;
const ll MOD=1000000007LL;
//ll MOD=1000000007LL;
//const ll MOD=998244353LL;
//const ll MOD=1777777777LL;
//const ll MAX_V=114514LL;
//const ll MAX = 500010LL;
const ll mod=MOD;
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
vl dx={0,0,1,-1};
vl dy={1,-1,0,0};

//素因数分解O(√n)
map<ll,ll>prime_factor(ll n){
  map<ll,ll>res;
  for(ll i=2;i*i<=n;i++){
    while(n%i==0){
      res[i]++;
      n/=i;
    }
  }
  if(n!=1)res[n]=1;
  return res;
}

const ll MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
//finvが階乗の逆元

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modpow(ll a, ll n, ll mod) {
  
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}



int main(){
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(10);
  /*--------------------------------*/
  ll n;cin>>n;
  cout<<10-n/200<<endl;
}