Problem:
Score: 200 points

Problem Statement

M-kun has the following three cards:

A red card with the integer A.

A green card with the integer B.

A blue card with the integer C.

He is a genius magician who can do the following operation at most K times:

Choose one of the three cards and multiply the written integer by 2.

His magic is successful if both of the following conditions are satisfied after the operations:

The integer on the green card is strictly greater than the integer on the red card.

The integer on the blue card is strictly greater than the integer on the green card.

Determine whether the magic can be successful.

Constraints

1 \leq A, B, C \leq 7

1 \leq K \leq 7

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C
K

Output

If the magic can be successful, print Yes; otherwise, print No.

Sample Input 1

7 2 5
3

Sample Output 1

Yes

The magic will be successful if, for example, he does the following operations:

First, choose the blue card. The integers on the red, green, and blue cards are now 7, 2, and 10, respectively.

Second, choose the green card. The integers on the red, green, and blue cards are now 7, 4, and 10, respectively.

Third, choose the green card. The integers on the red, green, and blue cards are now 7, 8, and 10, respectively.

Sample Input 2

7 4 2
3

Sample Output 2

No

He has no way to succeed in the magic with at most three operations.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define int int64_t
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = (int)(n) - 1; (i) >= (int)(m); -- (i))
#define REPP(i, n) for (int i = 1; (i) < (int)(n+1); ++ (i))
#define REPR(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define FORR(i, m, n) for(int i=(m)-1; i>=(n); i--)
#define ALL(v) v.begin(), v.end()
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())
#define SORTR(c) sort((c).begin(),(c).end(),greater<int>());
#define BIT(n) (1LL<<(n))
#define PCNT(x) __builtin_popcountll(x)
#define GCD(x,y) __gcd(x,y)
#define LCM(x,y) x/__gcd(x,y)*y
#define PB push_back
#define MP make_pair
#define endl "\n"
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
using namespace std;
using Graph = vector<vector<int> >;
using WeightedGraph = vector<vector<pair<int, int> > >;
typedef long double LD;
typedef vector<int> VI;
typedef vector<vector<int> > VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef vector<PII> VPII;
const int INF = 2000000000;
const long double PI = acosl(-1);
const int MOD = 1000000007;
void print() {cout << endl;}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head; if (sizeof...(tail) != 0) cout << " ";print(forward<Tail>(tail)...);}
template <class T> void print(vector<T> &vec) {for (auto& a : vec) {cout << a; if (&a != &vec.back()) cout << " ";} cout << endl;}
template <class T> void print(vector<vector<T>> &df) { for (auto& vec : df) { print(vec);}}


signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cout << fixed << setprecision(10);
    int ans = 0;
    
    int A, B, C, K;
    cin >> A >> B >> C >> K;

    while (A >= B) {
        K--;
        B *= 2;
    }
    while (B >= C) {
        K--;
        C *= 2;
    }
    if (K >= 0) yes;
    else no;
    
    
    
    return 0;
}

/*
cd ~/Desktop/Programming/AtCoder/contest/m-solutions2020/m_solutions2020_b && sh shellscripts/test_cpp.sh
cd ~/Desktop/Programming/AtCoder/contest/m-solutions2020/m_solutions2020_b && sh shellscripts/etest_cpp.sh
cd ~/Desktop/Programming/AtCoder/contest/m-solutions2020/m_solutions2020_b && sh shellscripts/submit_cpp.sh
*/