Codeforces Round #355 (Div. 2) A

A. Vanya and Fence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai.

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 ai (1 ≤ ai ≤ 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 to2 + 2 + 2 + 2 + 2 + 1 = 11.

 题意：输入n h  n个数 大于h的自增2 小于1的自增1

 题解：水

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<cmath>
#define ll __int64 
#define pi acos(-1.0)
#define mod 1000000007
using namespace std;
int n,h;
int exm;
int main()
{
    scanf("%d %d",&n,&h);
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&exm);
        if(exm>h)
        ans+=2;
        else
        ans+=1;
    }
    cout<<ans<<endl;
    return 0;
}