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>
using namespace std;

typedef long long ll;
typedef long double ld;

#define pb push_back
#define sort(s) sort(s.begin(),s.end())
#define reverse(s) reverse(s.begin(),s.end())
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)

const ll mod = 1e9+7;

//最大公約数
ll gcd(ll a, ll b) {
    return b ? gcd(b, a%b) : a;
}

//最小公倍数
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

//素数判定
bool isPrime(ll x){
    ll i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

//桁和
int digsum(ll n) {
    int res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

//桁数
int dignum(ll n) {
    int res = 0;
    while(n > 0) {
        res++;
        n /= 10;
    }
    return res;
}

//文字列中の特定の文字カウント
ll stringcount(string s,char c){
    return count(s.cbegin(),s.cend(),c);
}

//階乗
ll ka(ll i){
    ll res=1;
    while(i>0){
        res=res*i;
        i--;
    }
    return res;
}
 
//ncr
ll ncr(ll x,ll y){
    ll a = x;
    ll b = 1;
    if(y>(x/2)){
        y=x-y;
    }
    for(ll i=1;i<y;i++){
        a*=x-i;
        b*=i+1;
    }
    return a/b;
}

//フィボナッチ数列
ll f(ll x){
    ll dp[x+1];
    dp[1]=1;
    dp[2]=1;
    for(ll i=3;i<=x;i++){
        dp[i]=dp[i-1]+dp[i-2];
    }
    return dp[x];
}

//char->int
int ctoi(char c){
    return c-'0';
}

vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}



int main(){
    int a,b,c,k;
    cin >> a >> b >> c >> k;

    for(int i=0;i<k;i++){
        if(a>=b){
            b = b*2;
        }else{
            c = c*2;
        }

        if(b>a&&c>b){
            cout << "Yes" << endl;
            return 0;
        }
    }

    cout << "No" << endl;
}