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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>

#define SIZE 100005
#define PI acosl(-1) //3.14159265358979323846264338327950L 
#define rep(i, N) for (i = 0; i < N; i++) //制御変数iを用いてN回転
#define array(N, t) (t*)calloc(N, sizeof(t)) //t型N要素の1次元配列を動的確保後0クリア(freeを忘れずに)
#define zero(a); {int iter; rep(iter, sizeof(a)/sizeof((a)[0])) (a)[iter]=0;} //1次元配列aのゼロ初期化?
#define Sort(a, t) qsort(a, sizeof(a)/sizeof((a)[0]), sizeof(t), comp) //compを使って1次元配列をソート

typedef long long ll; typedef long double ld;
typedef struct node { int data; struct node *next; } Node;

Node *list[SIZE];
int queue[SIZE]; int q_head = 0, q_tail = 0;

void print_log(); //とりあえず文字を出力してどこまで実行できているか確認(for debug)
void printa_int(int *a, int size); //int型の1次元配列aを出力(for debug)
void printl_int(int size); //リンクリストを出力(for debug)
int comp(const void *a, const void *b); //昇順並び替えに使う比較関数
Node* add_node(int i, int data); void free_list(Node* head);
int queue_empty(); int queue_full();
void enqueue(int data); int dequeue();
ll* factorize(ll N); //Nを素因数分解、得られる配列は素因数,指数,素因数,…となっている(添字はintで指定する必要あり?)
void check_factorization(ll *fact); //素因数分解を表示(for debug)

int main() {
    int A, B, C, K, count = 0;
    scanf("%d %d %d %d", &A, &B, &C, &K);
    while (A >= B) {
        B *= 2;
        count++;
    }
    while (B >= C) {
        C *= 2;
        count++;
    }
    if (count <= K) printf("Yes\n");
    else printf("No\n");
    return 0;
}

void print_log() {
    printf("=== print log ===\n");
}
void printa_int(int *a, int size) {
    int i;
    printf("[");
    for (i = 0; i < size; i++)
        printf("%d, ", a[i]);
    printf("]\n");
}
void printl_int(int size) {
    int i;
    Node *t;
    rep(i, size) {
        printf("[%d] ", i);
        for (t = list[i]; t; t = t->next) printf("%d, ", t->data);
        printf("\n");
    }
}
int comp(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}
Node* add_node(int i, int data) {
    if (!list[i]) {
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
    } else {
        Node *t;
        t = list[i];
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
        list[i]->next = t;
    }
    return list[i];
}
void free_list(Node* head) {
    Node *prev = NULL, *t = head;
    while (t) {
        prev = t;
        t = t->next;
        free(prev);
    }
}
int queue_empty() {
    return q_head == q_tail;
}
int queue_full() {
    return (q_head - 1) % SIZE == q_tail % SIZE;
}
void enqueue(int data) {
    if (queue_full()) {
        printf("full!\n");
        return;
    }
    queue[q_tail++] = data;
    q_tail %= SIZE;
}
int dequeue() {
    if (queue_empty()) {
        printf("empty!\n");
        return -444444;
    }
    int data = queue[q_head++];
    q_head %= SIZE;
    return data;
}
ll* factorize(ll N) {
    ll *res = array(((sqrt(N) + 1) * 2), ll);
    ll i, x = N;
    int count = 0;
    for (i = 2; i * i <= N; i++) {
        ll e = 0;
        while (x % i == 0) {
            x /= i;
            e++;
        }
        if (e > 0) {
            res[count * 2] = i;
            res[count * 2 + 1] = e;
            count++;
        }
    }
    if (x > 1) {
        res[count * 2] = x;
        res[count * 2 + 1] = 1;
    }
    return res;
}
void check_factorization(ll *fact) {
    int i = 1;
    printf("%lld^%lld", fact[0], fact[1]);
    while (fact[i * 2] > 0) {
        printf(" * %lld^%lld", fact[i * 2], fact[i * 2 + 1]);
        i++;
    }
    printf("\n");
}