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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

public class ABC156_A
{

    public static void Main()
    {
        int a = 0, b = 0;
        InLine(ref a, ref b);
        if (a < 10) b += (100 * (10 - a));
        else Out(b);
    }

    //Template
    private static object[] nullobjs = null;
    private static object nullobj = null;

    public static int ConvertInt(char c)
    {
        return c;
    }

    public static void InArgs<A>(int count, ref A[] var)
    {
        InArgs(count, ref var, ref nullobjs, ref nullobjs);
    }

    public static void InArgs<A, B>(int count, ref A[] var1, ref B[] var2)
    {
        InArgs(count, ref var1, ref var2, ref nullobjs);
    }

    public static void InArgs<A, B, C>(int count, ref A[] var1, ref B[] var2, ref C[] var3)
    {
        A a = default(A);
        B b = default(B);
        C c = default(C);
        for (int i = 0; i < count; i++)
        {
            InLine(ref a, ref b, ref c);
            var1[i] = a;
            if (var2 != null) var2[i] = b;
            if (var3 != null) var3[i] = c;
        }
    }

    public static void InLine<A>(ref A var)
    {
        InLine(ref var, ref nullobj, ref nullobj);
    }

    public static void InLine<A, B>(ref A var1, ref B var2)
    {
        InLine(ref var1, ref var2, ref nullobj);
    }

    public static void InLine<A, B, C>(ref A var1, ref B var2, ref C var3)
    {
        object[] o = new object[] { var1, var2, var3 };
        Type[] t = new Type[] { typeof(A), typeof(B), typeof(C) };
        if (var3 == null)
        {
            if (var2 == null) o = new object[] { var1 };
            else o = new object[] { var1, var2 };
        }

        string[] s = Console.ReadLine().Split(' ');
        for (int i = 0; i < s.Length; i++)
        {
            if (typeof(int) == t[i]) o[i] = int.Parse(s[i]);
            else if (typeof(long) == t[i]) o[i] = long.Parse(s[i]);
            else if (typeof(double) == t[i]) o[i] = double.Parse(s[i]);
            else o[i] = s[i];
        }
        var1 = (A)o[0];
        if (var2 != null) var2 = (B)o[1];
        if (var3 != null) var3 = (C)o[2];
    }

    public static string JoinString(object[] obj, string sep)
    {
        return string.Join(sep, obj.Select(i => i.ToString()).ToArray());
    }

    public static void Out(object o)
    {
        Console.WriteLine(o);
    }

    public static void OutYes()
    {
        Console.WriteLine("Yes");
    }

    public static void OutNo()
    {
        Console.WriteLine("No");
    }
    public static void OutYES()
    {
        Console.WriteLine("YES");
    }

    public static void OutNO()
    {
        Console.WriteLine("NO");
    }

    private static Dictionary<A, B> PutDict<A, B>(A[] key, B[] value)
    {
        if (key.Length != value.Length) return null;
        var dic = new Dictionary<A, B>();
        for (int i = 0; i < key.Length; i++)
        {
            if (!dic.ContainsKey(key[i])) dic.Add(key[i], value[i]);
            else dic[key[i]] = value[i];
        }
        return dic;
    }
}