D - Vanya and Fence

Problem description

Vanya and his friends are walking along the fence of height h and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed h. If the height of some person is greater than h he can bend down and then he surely won't be noticed by the guard. The height of the i-th person is equal to a_i.

Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?

Input

The first line of the input contains two integers n and h (1 ≤ n ≤ 1000, 1 ≤ h ≤ 1000) — the number of friends and the height of the fence, respectively.

The second line contains n integers a_i (1 ≤ a_i ≤ 2h), the i-th of them is equal to the height of the i-th person.

Output

Print a single integer — the minimum possible valid width of the road.

Examples

Input

3 7
4 5 14

Output

4

Input

6 1
1 1 1 1 1 1

Output

6

Input

6 5
7 6 8 9 10 5

Output

11

Note

In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.

In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.

In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.

解题思路：如果某个人的高度大于篱笆的高度h，那么这个人必须弯着身子走路才不会被警卫发现，其宽度为2；否则其宽度为1，计算n个人（人与人紧凑着）走路排成一行共有多宽，水过！

AC代码：

 #include<bits/stdc++.h>
 using namespace std;
 int main(){
     int n,h,a,sum=;
     cin>>n>>h;
     while(n--){
         cin>>a;
         if(a>h)sum+=;
         else sum++;
     }
     cout<<sum<<endl;
     return ;
 }