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)

#pragma GCC optimize("Ofast","unroll-loops")
#pragma GCC push_options
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#include <xmmintrin.h>
#include <immintrin.h>
using namespace::std;
__attribute__((constructor))void init(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
#include<ext/pb_ds/tag_and_trait.hpp>
template<typename T>using pbds=__gnu_pbds::tree<T,__gnu_pbds::null_type,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T>using pbds_map=__gnu_pbds::tree<T,T,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T,typename E>using hash_map=__gnu_pbds::gp_hash_table<T,E>;
template<typename T>using pqueue =__gnu_pbds::priority_queue<T, less<T>,__gnu_pbds::rc_binomial_heap_tag>;
typedef long long lint;
#define INF (1LL<<60)
#define IINF (1<<30)
#define EPS (1e-10)
typedef vector<lint> vec;
typedef vector<vector<lint>> mat;
template<typename T>inline void numout(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i<INF/2?i:"INF";f=1;}cout<<endl;}
template<typename T>inline void numout2(T t){for(auto i:t)numout(i);}
template<typename T>inline void output(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i;f=1;}cout<<endl;}
template<typename T>inline void output2(T t){for(auto i:t)output(i);}
template<typename T>inline void _output(T t){bool f=0;for(int i=0;i<t.size();i++){cout<<f?"":" "<<t[i];f=1;}cout<<endl;}
template<typename T>inline void _output2(T t){for(int i=0;i<t.size();i++)output(t[i]);}
#define rep(i,n) for(int i=0;i<int(n);++i)
#define repi(i,a,b) for(int i=int(a);i<(int)(b);++i)
#define rrep(i,n) for(int i=int(n)-1;i>=0;--i)
#define rrepi(i,a,b) for(int i=int(b)-1;i>=int(a);--i)
#define irep(i,n) for(int i=int(n);;++i)
#define all(n) begin(n),end(n)
inline auto gcd(auto A,auto B){return B?gcd(B,A%B):A;}
inline auto lcm(auto A,auto B){return A/gcd(A,B)*B;}
inline bool chmin(auto& s,const auto& t){bool res=s>t;s=min(s,t);return res;}
inline bool chmax(auto& s,const auto& t){bool res=s<t;s=max(s,t);return res;}
const vector<lint> dx={-1,1,0,0,1,1,-1,-1};
const vector<lint> dy={0,0,-1,1,1,-1,1,-1};
#define SUM(v) accumulate(all(v),0LL)
auto call=[](auto f,auto... args){return f(f,args...);};

int main(){
    lint k,x;
    cin>>k>>x;
    cout<<(k*500>=x?"Yes":"No")<<endl;
}