Given a sequence of positive integers and another positive integer p. The sequence is said to be a perfect sequence if M≤m×p where M and m are the maximum and minimum numbers in the sequence, respectively.

Now given a sequence and a parameter p, you are supposed to find from the sequence as many numbers as possible to form a perfect subsequence.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (≤10​5​​) is the number of integers in the sequence, and p (≤10​9​​) is the parameter. In the second line there are N positive integers, each is no greater than 10​9​​.

Output Specification:

For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.

Sample Input:

10 8
2 3 20 4 5 1 6 7 8 9

Sample Output:

8


注意点：1.int*int 的范围必须用long long;
　　　　2.if(a[n-1]<=x) return n;

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn =100010;
 5 
 6 int a[maxn];
 7 
 8 int n,p;
 9 
10 int binary_Search(int i,long long x){
11     
12     if(a[n-1]<=x)
13         return n;
14     
15     int left=i+1,right=n-1;
16     
17     while(left<right){
18         int mid = (left+right)/2;
19         
20         if(a[mid]>x){
21             right =mid;
22         }else{
23             left= mid+1;
24         }
25     }
26     
27     return left;
28     
29     
30 }
31 
32 
33 int main(){
34 
35     cin>>n>>p;
36     
37     for(int i=0;i<n;i++)
38         cin>>a[i];
39         
40     sort(a,a+n);
41     
42     int ans=1;
43     
44     for(int i=0;i<n;i++){
45         
46     //    int j=binary_Search(i,(long long)a[i]*p);
47         
48         int j=upper_bound(a+i+1,a+n,(long long)a[i]*p)-a;
49             
50         ans=max(ans,j-i);
51         
52     //    cout<<j<<" "<<i<<endl;
53     }
54     
55 
56     
57     cout<<ans<<endl;
58     
59     return 0;
60     
61 }