
// Problem: A - ABC Preparation
// Contest: AtCoder - AtCoder Beginner Contest 185
// URL: https://atcoder.jp/contests/abc185/tasks/abc185_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update

using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
ordered_set;	

struct splitmix64_hash {
	static uint64_t splitmix64(uint64_t x) {
		// http://xorshift.di.unimi.it/splitmix64.c
		x += 0x9e3779b97f4a7c15;
		x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
		x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
		return x ^ (x >> 31);
	}
 
	size_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = std::chrono::steady_clock::now().time_since_epoch().count();
		return splitmix64(x + FIXED_RANDOM);
	}
};
 
template <typename K, typename V, typename Hash = splitmix64_hash>
using hash_map = __gnu_pbds::gp_hash_table<K, V, Hash>;
 
template <typename K, typename Hash = splitmix64_hash>
using hash_set = hash_map<K, __gnu_pbds::null_type, Hash>;

#define FOR(i,a,b) for(int i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(int i = (a); i >= (b); --i)
#define RI(i,n) FOR(i,1,(n))
#define REP(i,n) FOR(i,0,(n)-1)
#define mini(a,b) a=min(a,b)
#define maxi(a,b) a=max(a,b)
#define pb push_back
#define st first
#define nd second
#define sz(w) (int) w.size()
typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<pii, int> para;
const ll inf = 1e18 + 7;
const ll maxN = 1e6 + 5;
const ll MOD = 1e9 + 7;

// sprawdz MODULO!
int main() {
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int a, b, c, d;
	cin >> a >> b >> c >> d;
	cout << min(min(a, b), min(c, d));
	return 0;
}
 

#include<iostream>
using namespace std;

int main(){
    
    int V, T, S, D;
    
    cin >> V >> T >> S >> D;
    
    if(D / V >= T && S >= ((D + (V-1))/ V)){
        cout << "No" << endl;
    }else{
        cout << "Yes" << endl;
    }
    
    return 0;
}