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>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
 
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("-ffloat-store")
 
#define int  long long
#define rep(i,begin,end) for(__typeof(end)i=(begin)-((begin)>(end));i!=(end)-((begin)>(end));i+=1-2*((begin)>(end)))
#define F first
#define S second
#define sz(x) ((int)x.size())
#define pb push_back
#define pf push_front
#define eb emplace_back
#define all(v) (v).begin(), (v).end()
#define mod 1000000007
#define what_is(x) cerr << #x << " is " << x <<"\n";
#define sortA(v) sort(v.begin(), v.end())
#define sortD(v) sort(v.rbegin(), v.rend())
#define PI 3.14159265358979323846
#define vout(a) for(auto x:a) cout<<x<<" ";
#define INF 10000000000000000 //1e16
 
// #define ordered_set tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
 
 
// using namespace __gnu_pbds;
using namespace std;
 
using ll  = long long;
using vi  = vector<int>;
using pii = pair<int,int>;
using vii = vector<pii>;
using mi  = multiset<int>;
using mii = multiset<pii>;
 
int lcm(int a,int b) {return (a*(b/__gcd(a,b)));}


 
void solve()
{
    int n,m,t,i,k,r,l,u,v,w,j,s,sum=0,pos,flag=0,count=0;
    string S,T,U; 
    int a,b,c;
    cin>>a>>b>>c>>k;
    if(b<=a)
    {
        while(k--)
        {
            b*=2;
            if(b>a) break;
        }
    }
    if(b<=a)
    {
        cout<<"No";
        return;
    }
    if(c<=b)
    {
        while(k--)
        {
            c*=2;
            if(c>b) break;
        }
    }
    if(c<=b)
    {
        cout<<"No";
        return;
    }
    cout<<"Yes";
}
 
 
signed main()
{
    #ifndef ONLINE_JUDGE 
       freopen("input.txt","r",stdin);
       freopen("output.txt","w",stdout);
    #endif
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    // cin>>t;
    while(t--)
    {
        solve();
        // if(t)cout<<"\n";
    }
    return 0;
}