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: Accepted)

#include<bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define dofloat cout<<fixed<<setprecision(8)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define bitcount __builtin_popcount
#define all(vec) vec.begin(),vec.end()
#define rall(vec) vec.rbegin(),vec.rend()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<long long> vl;
typedef pair<long long,long long> pll;
typedef vector<pair<long long,long long> > vll;
typedef vector<pair<int,int> > vii;
typedef vector<int> vi;
typedef pair<int,int> ii;
const long long MOD=1000000007;
const long long MAX=100005;
const long double PI=3.14159265359;
const long double G=9.807;
const long long INF=1e18;
const long double EPS=1e-6;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
bool isprime(ll a){
  if(a==2)return 1;
  if(!(a&1))return 0;
  for(ll i=3;i*i<=a;i+=2)
    if(a%i==0)return 0;
  return 1;
}
ll mpow(ll base,ll exponent,ll modulus){
  if(modulus==1)return 0;
  long long result = 1;
  base=base%modulus;
  while(exponent){
    if(exponent%2 == 1)
    result=(result*base)%modulus;
    exponent=exponent>>1;
    base=(base*base)%modulus;
  }
  return result;
}
ll minv(ll a,ll mod){
  ll _gcd=gcd(a,mod);
  assert(_gcd==1);
  return mpow(a,mod-2,mod);
}


/*
ll ncr(ll N,ll K){
  if(N<K)return 0;
  if(K==0)return 1;
  if(N==0)return 0;

  ll den=1;
  for(ll i=1;i<=K;i++)den*=i;
  ll num=1;
  while(K--){
    num*=N;
    if(num%den==0){
      num/=den;
      den=1;
    }
    N--;
  }

  return num;
}
*/
ll n,r;
int main(){FAST;
cin>>n>>r;
if(n<10){
  cout<<r+100*(10-n);
}
else{
  cout<<r;
}
return 0;
}