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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <cstring>
#include <climits>
using namespace std;
#define ll long long
#define FOR(i,n) for(int i=0;i<n;i++)
#define sz(x) (int)(x.size())
#define all(x) (x).begin(),(x).end()
template<typename tp> void read(tp &x) {cin >> x;}
template<typename tp, typename... Args> void read(tp& x, Args&... args) {read(x);read(args...);}
const int mxN = (int)1e5,INF = (int)1e9,MOD = (int)1e9+7;

void solve()
{
	int k,x;
	read(k,x);
	cout << (k*500>=x?"Yes":"No");
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int test = 1;
	//read(test);
	while(test--)
		solve();
	return 0;
}