codeforces——677A——Vanya and Fence

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 exceedh. If the height of some person is greater thanh he can bend down and then he surely won't be noticed by the guard. The height of thei-th person is equal toa_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 andh (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), thei-th of them is equal to the height of thei-th person.

Output

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

Examples

Input

3 74 5 14

Output

4

Input

6 11 1 1 1 1 1

Output

6

Input

6 57 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 to1 + 1 + 2 = 4.

In the second sample, all friends are short enough and no one has to bend, so the width1 + 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.

输入n,m，然后输入n个数，输入的数小于等于m，则值为1，反之为2，求总值。

水

#include <iostream>#include<cstring>#include<string>#include<stdio.h>#include<algorithm>#include<set>using namespace std;int main(){    int n,m;    while(cin>>n>>m!=NULL)    {        int t=n,ans=0;        while(t--)        {            int t2;            cin>>t2;            if(t2<=m)                ans++;        }        cout<<(n-ans)*2+ans<<endl;    }    return 0;}