#include <bits/stdc++.h>
using namespace std;

#define pii  pair<int, int>
#define vi vector<int >
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define sz(x) (x).size()
#define all(x) (x).begin(), (x).end() //Forward traversal
#define rall(x) (x).rbegin, (x).rend() //reverse traversal
#define endl "\n"

#define ll long long
#define ld long double

const ld EPS = 10e-9;
const int INF = 1e9 + 5;
const long long INFLL = (ll)INF * (ll)INF;
const ll MOD = 1e9 + 7;
const ld PI = 3.14159265358979323846;

struct HASH{
  size_t operator()(const pair<int,int>&x)const{
    return hash<long long>()(((long long)x.first)^(((long long)x.second)<<32));
  }
};

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "[";
    for (int i = 0; i < (int)v.size(); ++i) {
        os << v[i];
        if (i != (int)v.size() - 1)
            os << ", ";
    }
    os << "]\n";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it;
        if (it != *v.rbegin())
            os << ", ";
    }
    os << "}\n";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const unordered_set<T>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it;
        if (it != *v.rbegin())
            os << ", ";
    }
    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it.first << " : "
           << it.second;

        if( it != *v.rbegin())
            os  << ", ";
    }

    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const unordered_map<T, S>& v)
{
    os << "{";
    for (const auto& it : v) {
        os << it.first << " : "
           << it.second;

        if( it != *v.rbegin())
            os  << ", ";
    }

    os << "}\n";
    return os;
}

template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v)
{
    os << "(";
    os << v.first << ", "
       << v.second << ")";
    os << "\n";
    return os;
}

#define DEBUG

#ifdef DEBUG
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define debug(...)
#endif


// increase all elements by x, query min and max elements log(n)
//
struct Offset {
    set<ll> s;
    ll delta;

    ll getMinimum() {
        return (*s.begin()) + delta;
    }

    ll getMaximum() {
        return (*s.rbegin()) + delta;
    }

    void increase(ll x) {
        delta += x;
    }

    void add(ll x) {
        s.insert(x-delta);
    }

    void remove(ll x) {
        s.erase(x-delta);
    }

    bool contains(ll x) {
        return s.count(x-delta) != 0;
    }
};



// __gcd(x, y)

// idea behind binary search: there is monotonically increasing predicate and range [l, r]. Find
// x in range such that x is smallest value for which predicate holds true.
//            int lo = 1, hi = 1e9;
//            while(lo < hi)
//            {
//                ll mid = (lo+hi) / 2;
//                if(predicate(mid))
//                {
//                    hi = mid;
//                }
//                else
//                {
//                    lo = mid+1;
//                }
//            }
//            assert(lo == hi);

// OR: given monotonically decreasing predicate, find x in range such that x is largest value for
// which predicate holds true.

//int x = 0;
//for(int s = max_step; s >= 1; s /= 2)
//{
//    while(predicate(x+s)) x += s;
//}
// after this, x is the largest value for which predicate is true

// How to calculate mid element:
// floor((l+r) / 2) might overflow, use l + floor((r-l)/2)



//ll l = 0, r = s.size()-1;
//ll ans = -1;
//while(l < r)
//{
//    ll mid = l + (r-l) / 2;
//    if(P(mid)) {
//        r = mid;
//        ans = mid;
//    }
//    else
//    {
//        l = mid + 1;
//    }
//}
//
// AND
//
//ll l = 0, r = s.size()-1;
//ll ans = -1;
//while(l <= r)
//{
//    ll mid = l + (r-l) / 2;
//    if(P(mid)) {
//        r = mid-1;
//        ans = mid;
//    }
//    else
//    {
//        l = mid + 1;
//    }
//}
//
// ARE EQUIVALENT

//__builtin_popcount(ll) - returns number of set bits
//__builtin_clz(ll) - count leading zeroes
//__builtin_ctz(ll) - count trailing zeroes


// iterate over all subsets of x (x is a set formed like x = (1<<3)|(x<<6)|...
//int b = 0;
//do {
// process subset b
//} while (b=(b-x)&x);


// iterate over all subsets from 0 to 2^n-1
//for (int b = 0; b < (1<<n); b++) {
//vector<int> subset;
//for (int i = 0; i < n; i++) {
//if (b&(1<<i)) subset.push_back(i);
//}
//}

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

    int a, b, c, k;
    cin >> a >> b >> c >> k;
    if(a < b && b < c) {
        cout << "Yes" << endl;
    }
    else {
        if(a >= b*pow(2, k)) {
            cout << "No" << endl;
        }
        else {
            int p = 0;
            while(b <= a) {
                p++;
                b *= 2;
            }
            int left = k - p;
            if(b >= c*pow(2, left)) {
                cout << "No" << endl;
            }
            else {
                cout << "Yes" << endl;
            }
        }
    }

    return 0;
}


Measured memory_bytes: 4108