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 _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

// repetition
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

// container util
#define all(x) (x).begin(), (x).end()

// debug
#define dump(x) cerr << #x << " = " << (x) << endl;
#define debug(x)                                         \
  cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" \
       << " " << __FILE__ << endl;

// typedef
typedef long long lint;
typedef unsigned long long ull;
typedef complex<long double> Complex;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

// constant
const int MOD = (int)1e9 + 7;
const int INF = (int)1e18;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};

// conversion
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
//

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int k, x;
  cin >> k >> x;
  if (x <= k * 500) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }

  return 0;
}