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
// #define _USE_MATH_DEFINES	// M_PI=3.1415...
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <functional>
#include <locale>
#include <cctype>
#include <sstream>
#include <iomanip>	// 20桁出力 cout << setprecision(20) << double;
#include <queue>
#include <bitset>	// bitset<8> bs1(131uL); // 10000011 bs1[0]は1 01stringからビット集合生成可
#include <numeric>
#include <random>
#include <climits>  // A/Bを計算するときA==LLONG_MIN && B==-1のみ(1ull<<63)で代用
using namespace std;

typedef long long LL;
typedef pair<LL, LL> P;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VL;
typedef vector<VL> VVL;				// VVL v(100,VL(100,0));
typedef vector<pair<LL, LL> > VP;
typedef vector<string> VS;
typedef map<int, int> MAPII;
typedef multimap<int, char, greater<int> > MuMAPIC;
typedef multimap<int, string, greater<int> > MuMIS;
typedef pair<LL, pair<P, P> > PP;

#define INF 999999999999999997		// 少し加算したらオーバーフローする
#define MP make_pair
#define FAST_IO  cin.tie(0); ios::sync_with_stdio(false);
#define FOR(i,a,b) for(LL i=(a);i<(b);i++)
#define FOR_REV(i,a,b) for(int i=(a);i>=(b);i--)
#define FOR_ITR(d) for(auto itr=d.begin(),d_end=d.end();itr!=d_end;++itr)	// C++11
#define O(s) cout<<s<<endl;
#define SORTVL(v) sort(v.begin(), v.end());
#define SORTVL_GR(v) sort(v.begin(), v.end(), greater<LL>());

#define DUMP_VVI(b) if(dbgF){FOR(i,0,b.size()){FOR(j,0,b[i].size())printf("%d ",b[i][j]);puts("");}}
#define D_OUT(str,value) if(dbgF){cout<<str<<" : "<<value<<endl;}

template<class T>T IN() { T d; cin >> d; return d; }

LL gcd(LL a, LL b) { return (b > 0) ? gcd(b, a % b) : a; } // 最大公約数（Greatest Common Divisor）
LL lcm(LL a, LL b) { return a / gcd(a, b) * b; } // 最小公倍数（Least Common Multiple）
bool uruu(LL Y) { return (((Y % 4 == 0 && Y % 100 != 0) || Y % 400 == 0) ? true : false); } // うるう年判定
string Replace(string S, string from, string to) { auto p = S.find(from); while (p != string::npos) { S.replace(p, from.size(), to); p = S.find(from, p + to.size()); }return S; }
// 桁和
LL Ketawa_s(string s) { LL a = 0; FOR(i, 0, s.length())a += s[i] - '0'; return a; }
// 末尾数字で切り上げ
LL CeilLastNumber(LL n) { return ((n + (10 - 1)) / 10) * 10; }
//二項係数 nCk mod pを求める
const LL MAX = 510000, p = 998244353; LL fac[MAX], finv[MAX], inv[MAX];
/* 前処理 */void COM_init() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; FOR(i, 2, MAX) { fac[i] = fac[i - 1] * i % p; inv[i] = p - inv[p % i] * (p / i) % p; finv[i] = finv[i - 1] * inv[i] % p; } }
/* 主処理 */LL COM(int n, int k) { if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % p) % p; }

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };
int dxx[8] = { 0,1,1,1,0,-1,-1,-1 };
int dyy[8] = { 1,1,0,-1,-1,-1,0,1 };

// -------------------------------------------------------------------------------

int main() {
	FAST_IO;
	bool dbgF = true; // for D_OUT(str, value)  ... cout<< str <<" : "<< value <<endl;

	LL K, X;
	cin >> K >> X;

	if (500 * K >= X) {
		O("Yes");
	}
	else {
		O("No");
	}


	return 0;
}