Problem:
Score : 100 points

Problem Statement

Takahashi is a member of a programming competition site, ButCoder.

Each member of ButCoder is assigned two values: Inner Rating and Displayed Rating.

The Displayed Rating of a member is equal to their Inner Rating if the member has participated in 10 or more contests. Otherwise, the Displayed Rating will be their Inner Rating minus 100 \times (10 - K) when the member has participated in K contests.

Takahashi has participated in N contests, and his Displayed Rating is R. Find his Inner Rating.

Constraints

All values in input are integers.

1 \leq N \leq 100

0 \leq R \leq 4111

Input

Input is given from Standard Input in the following format:

N R

Output

Print his Inner Rating.

Sample Input 1

2 2919

Sample Output 1

3719

Takahashi has participated in 2 contests, which is less than 10, so his Displayed Rating is his Inner Rating minus 100 \times (10 - 2) = 800.

Thus, Takahashi's Inner Rating is 2919 + 800 = 3719.

Sample Input 2

22 3051

Sample Output 2

3051

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
 
//#include <x86intrin.h>
#include <bits/stdc++.h>   
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace __gnu_pbds;
using namespace std;

template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
     
#define F first
#define S second           
#define lb lower_bound               
#define ub upper_bound
#define pb push_back
#define pf push_front    
#define ppb pop_back
#define mp make_pair                                     
#define bpp __builtin_popcountll                                                                                        
#define sqr(x) ((x) * (x)) 
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define in insert
#define ppf pop_front
#define endl '\n'
//#define int long long
 
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
 
const int mod = (int)1e9 + 7;
const int N = (int)3e5 + 123;
const ll inf = (ll)1e18 + 1;

const double pi = acos (-1.0);
const double eps = 1e-7;                  
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};

int n, r;

inline void boost () {                    
   ios_base :: sync_with_stdio (NULL);
   cin.tie (NULL), cout.tie (NULL);          
}                                                     

inline void Solve () {
   cin >> n >> r;
   if (n >= 10) cout << r, exit (0);
   cout << r - 100 * (10 - n);
}

int main () {                                       
// freopen (".in", "r", stdin);    
// freopen (".out", "w", stdout);              
   boost ();
   int tt = 1;
   //cin >> tt;   
   while (tt --) {
      Solve ();
   }                                               
   return 0;      
}