Problem:
Score : 100 points

Problem Statement

Takahashi has K 500-yen coins. (Yen is the currency of Japan.)
If these coins add up to X yen or more, print Yes; otherwise, print No.

Constraints

1 \leq K \leq 100

1 \leq X \leq 10^5

Input

Input is given from Standard Input in the following format:

K X

Output

If the coins add up to X yen or more, print Yes; otherwise, print No.

Sample Input 1

2 900

Sample Output 1

Yes

Two 500-yen coins add up to 1000 yen, which is not less than X = 900 yen.

Sample Input 2

1 501

Sample Output 2

No

One 500-yen coin is worth 500 yen, which is less than X = 501 yen.

Sample Input 3

4 2000

Sample Output 3

Yes

Four 500-yen coins add up to 2000 yen, which is not less than X = 2000 yen.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] str = Console.ReadLine().Split(' ');
            int K = int.Parse(str[0]);
            int X = int.Parse(str[1]);

            int sum = 500 * K;
            string ans;
            if (sum < X) ans = "No";
            else ans = "Yes";

            Console.WriteLine(ans);
        }
       
    }
}
/*
long a, b, c;
a = long.Parse(Console.ReadLine()); //標準入力
string[] str = Console.ReadLine().Split(' ');//2つ以上のスペース区切り入力の取得
b = long.Parse(str[0]);  //数値で受け取りたい場合は変換する
c = long.Parse(str[1]);
string S = Console.ReadLine();
char[] C = new char[N];
C = S.ToCharArray();
Console.Write("改行なし a = {0} b = {1} c = {2}", a, b, c)//改行なし出力
Console.WriteLine("改行あり a = {0} b = {1} c = {2}", a, b, c);//改行付き出力
Member = ReadLine().Split(' ').Select(long.Parse).ToArray();

string[] str = Console.ReadLine().Split(' ');
long[] a = new long[N];
for(long i = 0; i < N; i++) a[i] = long.Parse(str[i]);
 */
/*
      static long func(long n)
      {
          long d = 0;
          for (long i = 0; i < 10; i++)
          {
              if (n / longPow(10, i) == 0)
              {
                  d = i;
                  break;
              }
          }
          long ret = A * n + B * d;
          return ret;
      }
      static long longPow(long a, long n)
      {
          long b = 1;
          for (long i = 0; i < n; i++) b = b * a;
          return b;
      }
      */