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 int long long
#define endl '\n'
using namespace std;

typedef long long ll;
#define prr(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cout << *it << " = " << a << endl;
	err(++it, args...);
}

#define read(a) int a; cin >> a;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}

#define pri(a) cout << a << endl;
#define print(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(a) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;} cout << endl;
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll N = 2e4 + 1;

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	readb(n, r);
	if (n >= 10) cout << r;
	else cout << r + 100*(10 - n);
}