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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <numeric>

using namespace std;

typedef long long int lli;

int gcd(int a, int b) {
	return b != 0 ? gcd(b, a%b) : a;
}

int lcm(int a, int b) {
	return a * b / gcd(a, b);
}

#define N_MAX 310
#define MOD 1000000007

void solv()
{
	int K, X; scanf("%d%d", &K, &X);
	if (500 * K >= X) {
		printf("Yes");
	}
	else {
		printf("No");
	}

	return;
}

int main()
{
	FILE *fin = NULL, *fout = NULL;
	//fin = freopen("input.txt", "r", stdin);
	//fout = freopen("output.txt", "w", stdout);

	solv();

	//finalize
	if (NULL != fin) fclose(fin);
	if (NULL != fout) fclose(fout);

	return 0;
}