#include <bits/stdc++.h>

using namespace std;

class gfg {
    /* Function to get no of set bits in binary
representation of passed binary no. */
public:
    unsigned int countSetBits(int n)
    {
        unsigned int count = 0;
        while (n) {
            n &= (n - 1);
            count++;
        }
        return count;
    }
}g;

vector<long long >prime;

#define MOD 1000000007
#define primemod 998244353
#define ll long long int
#define maxv(all,A) *max_element(A.begin(),A.end())
#define minv(all,A) *min_element(A.begin(),A.end())
#define sumv(all,A,a) accumulate(A.begin(), A.end(),a)
#define FOR(i,n) for(long long int i=0;i<(n);i++)
#define NFOR(i,n) for(long long int i=n-1;i>=0;i--)
#define asort(all,A) sort((A).begin(),(A).end())
#define dsort(all,A,x) sort((A).begin(),(A).end(),greater<x>())
#define pb(a) push_back((a))
#define mk_pair(x,y,a,b) insert(pair<x,y >(a,b))
#define autv(x) vector<x>::iterator
#define autm(x,y) map<x,y>::iterator
#define auts(x) set<x,greater<x> >::iterator
#define ini_v(n,x,A) FOR(i,n){ x a; cin>>a; A.pb(a) ;}
#define ini_m(n,x,y,b,M) for(ll i=0;i!=n;i++){y a; cin>>a; M.mk_pair(x,y,b+i,a);}
#define IFOR(i,x,y,M) for(autm(x,y) i=M.begin(); i!=M.end();i++)
#define in(a) emplace(A.begin(),a)


ll fact(int n);
int hexadecimalToDecimal(char hexVal[]);
vector<vector<ll> > power(vector<vector<ll> >&a,ll n, ll p);
ll fact(ll n, ll p);
ll power(ll x, ll n, ll p);
ll nCr(ll n, ll r, ll p);
vector<vector<ll> > multiply(vector<vector<ll> > &a, vector<vector<ll> > &b, ll p);
void manipulated_seive(ll N);
int gcd(int a, int b);
void generatePrimeFactors();
ll calculateNoOFactors(ll n);
bool bipartite(vector<int>A[],int ini,int ver,int k);


vector<int>color(1000000,0);
//__int128_t a;





void solve()
{
  int a,b,c,d;
  cin>>a>>b>>c>>d;
  a=ceil((a*1.0)/d);
  c=ceil((c*1.0)/b);
  if(a>=c)
    cout<<"Yes";
  else
    cout<<"No";
  //cout<<a<<c;

}

void template1 ()
{
    ll t;
    cin>>t;
    while(t--)
    {
        solve();
        cout<<"\n";
    }
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
   //template1();
     solve();cout<<"\n";
    return 0;
}















ll fact(int n)
{
    // single line to find factorial
    return (n==1 || n==0) ? 1: n * fact(n - 1);
}


int hexadecimalToDecimal(char hexVal[])
{
    int len = strlen(hexVal);

    // Initializing base value to 1, i.e 16^0
    int base = 1;

    int dec_val = 0;

    // Extracting characters as digits from last character
    for (int i=len-1; i>=0; i--)
    {
        // if character lies in '0'-'9', converting
        // it to integral 0-9 by subtracting 48 from
        // ASCII value.
        if (hexVal[i]>='0' && hexVal[i]<='9')
        {
            dec_val += (hexVal[i] - 48)*base;

            // incrementing base by power
            base = base * 16;
        }

        // if character lies in 'A'-'F' , converting
        // it to integral 10 - 15 by subtracting 55
        // from ASCII value
        else if (hexVal[i]>='A' && hexVal[i]<='F')
        {
            dec_val += (hexVal[i] - 55)*base;

            // incrementing base by power
            base = base*16;
        }
    }

    return dec_val;
}


ll fact(ll n, ll p){
	ll ans = 1;
	for(ll i = 1; i <= n; i++){
		ans *= i;
		ans %= p;
	}
	return ans;
}
ll power(ll x, ll n, ll p)
{
	ll res = 1;
	if(n == 1) return x%p;
	if(n%2 == 1) res = x%p;

	ll temp = power(x,n/2,p);
	return res*((temp*temp)%p)%p;
}
ll inv(ll a, ll p)
{
	return power(a,p-2,p);
}
ll nCr(ll n, ll r, ll p)
{
	if(r > n) return 0;
	ll t1 = fact(n,p);
	ll t2 = inv(fact(r,p),p);
	ll t3 = inv(fact(n-r,p),p);

	return (((t1*t2)%p)*t3)%p;
}
vector<vector<ll> > multiply(vector<vector<ll> > &a, vector<vector<ll> > &b, ll p)
{
	int r1,c1,r2,c2;
	r1 = a.size(); c1 = a[0].size();
	r2 = b.size(); c2 = b[0].size();

	vector<vector<ll> > result(r1,vector<ll>(c2,0));

	if(c1 != r2) return result;

	for(int i = 0; i < r1; i++){
		for(int j = 0; j < c2; j++){
			for(int k = 0; k < c1; k++){
				result[i][j] += a[i][k]*b[k][j];
				result[i][j] %= p;
			}
		}
	}
	return result;
}

vector<vector<ll> > power(vector<vector<ll> >&a,ll n, ll p)
{
	int r = a.size();
	vector<vector<ll> > result(r,vector<ll>(r,0));
	for(int i = 0; i < r; i++) result[i][i] = 1;

	while(n > 0){
		if(n%2 == 1){
			result = multiply(result,a,p);
		}
		n /= 2;
		a = multiply(a,a,p);
	}
	return result;
}

const long long MAX_SIZE = 1000001;

// isPrime[] : isPrime[i] is true if number is prime
// prime[] : stores all prime number less than N
// SPF[] that store smallest prime factor of number
// [for Exp : smallest prime factor of '8' and '16'
// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ]
vector<long long >isprime(MAX_SIZE , true);

vector<long long >SPF(MAX_SIZE);

// function generate all prime number less then N in O(n)
void manipulated_seive(ll N)
{
    // 0 and 1 are not prime
    isprime[0] = isprime[1] = false ;

    // Fill rest of the entries
    for (long long int i=2; i<N ; i++)
    {
        // If isPrime[i] == True then i is
        // prime number
        if (isprime[i])
        {
            // put i into prime[] vector
            prime.push_back(i);

            // A prime number is its own smallest
            // prime factor
            SPF[i] = i;
        }

        // Remove all multiples of  i*prime[j] which are
        // not prime by making isPrime[i*prime[j]] = false
        // and put smallest prime factor of i*Prime[j] as prime[j]
        // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ]
        // so smallest prime factor of '10' is '2' that is prime[j] ]
        // this loop run only one time for number which are not prime
        for (long long int j=0;
             j < (int)prime.size() &&
             i*prime[j] < N && prime[j] <= SPF[i];
             j++)
        {
            isprime[i*prime[j]]=false;

            // put smallest prime factor of i*prime[j]
            SPF[i*prime[j]] = prime[j] ;
        }
    }
}

int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);

}


const int MAX = 1000001;

// array to store prime factors
int factor[MAX] = { 0 };

// function to generate all prime factors
// of numbers from 1 to 10^6
void generatePrimeFactors()
{
    factor[1] = 1;

    // Initializes all the positions with their value.
    for (int i = 2; i < MAX; i++)
        factor[i] = i;

    // Initializes all multiples of 2 with 2
    for (int i = 4; i < MAX; i += 2)
        factor[i] = 2;

    // A modified version of Sieve of Eratosthenes to
    // store the smallest prime factor that divides
    // every number.
    for (int i = 3; i * i < MAX; i++) {
        // check if it has no prime factor.
        if (factor[i] == i) {
            // Initializes of j starting from i*i
            for (int j = i * i; j < MAX; j += i) {
                // if it has no prime factor before, then
                // stores the smallest prime divisor
                if (factor[j] == j)
                    factor[j] = i;
            }
        }
    }
}

// function to calculate number of factors
ll calculateNoOFactors(ll n)
{
    if (n == 1)
        return 1;

    int ans = 1;

    // stores the smallest prime number
    // that divides n
    int dup = factor[n];

    // stores the count of number of times
    // a prime number divides n.
    int c = 1;

    // reduces to the next number after prime
    // factorization of n
    int j = n / factor[n];

    // false when prime factorization is done
    while (j != 1) {
        // if the same prime number is dividing n,
        // then we increase the count
        if (factor[j] == dup)
            c += 1;

        /* if its a new prime factor that is factorizing n,
           then we again set c=1 and change dup to the new
           prime factor, and apply the formula explained
           above. */
        else {
            dup = factor[j];
            ans = ans * (c + 1);
            c = 1;
        }

        // prime factorizes a number
        j = j / factor[j];
    }

    // for the last prime factor
    ans = ans * (c + 1);

    return ans;
}

bool bipartite(vector<int>A[],int ini,int ver,int k)
{
    vector<int>tem;
    for(int i=0;i<A[ini].size();i++)
    {
        tem.pb(A[ini][i]);
    }
    if(color[ini]!=0&&color[ini]!=k)
        return false;
    else
        color[ini]=k;
    for(int i=0;i<tem.size();i++)
    {
        if(color[tem[i]]!=0&&color[tem[i]]!=k*(-1))
            return false;
        else
        color[tem[i]]=k*(-1);
    }
    bool flag5=false;
    for(int i=0;i<ver;i++)
    {
        if(color[i]==0)
        {
            flag5=true;
            break;
        }
    }
    if(!flag5)
        return true;
    for(int i=0;i<tem.size();i++)
    {
        bipartite(A,tem[i],ver,k*(-1));
    }
    return true;



}





Measured memory_bytes: 22584