# using heap sort
import heapq
def cookies(k, nums):
    # Write your code here
    heapq.heapify(nums)
    count =0
    
    while(nums):
        x = heapq.heappop(nums)
        if x>=k:
            return count
        
        if not nums:
            return -1
        y = heapq.heappop(nums)
        
        #formula
        n = x + (2*y)
        heapq.heappush(nums, n)
        count+=1
        
    # loop exits without giving count
    return -1

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
    
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
        int k=scan.nextInt();
        PriorityQueue<Integer> queue=new PriorityQueue<Integer>();
        for(int i=0;i<n;i++)
            {
            queue.add(scan.nextInt());
        }
        int count=0;
        while(queue.peek()<k)
            {
            if(queue.size()>=2)
                {
            int x=queue.remove();
            int y=queue.remove();
            y=x+2*y;
            queue.add(y);
            count++;
            }
            else
                {
                count=-1;
                break;
            }
        }
        System.out.println(count);
    }
}

int cookies(int k, vector<int> arr) {
    priority_queue<int, vector<int>, greater<int>> pq(arr.begin(), arr.end());
    int count = 0;

    while (pq.size() > 1 && pq.top() < k)
    {
        int first = pq.top();
        pq.pop();
        int second = pq.top();
        pq.pop();

        pq.push(first + 2 * second);
        count++;
    }
    return (pq.top() >= k) ? count : -1;   
}

def cookies(k, A):
    import heapq

    heapq.heapify(A)
    count = 0
    while not all(i >= k for i in A): # until all cookies are >= k
        if len(A) == 1:
            return -1
        a, b = heapq.heappop(A), heapq.heappop(A)
        c = a + 2 * b
        heapq.heappush(A, c)
        count += 1
    return count