Codeforces Round #355 (Div. 2)-A

A. Vanya and Fence

题目连接：http://codeforces.com/contest/677/problem/A

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 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 to 2 + 2 + 2 + 2 + 2 + 1 = 11.

题意：有N人，还有一个高度，每个人不能超过这个高度H。每个人通过这个高度可以有2种方式，1是当高度低于H时，这个人的宽度占1个单位，2是当高度超过H时，这个人要弯腰通过，这个人占2个单位。问最少的宽度是多少？

思路：很明显，当a[i]<=H时，对答案的贡献是1，否则对答案的贡献是2.

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
typedef long long int LL;
#define INF 0x3f3f3f3f
const int MAXN = 1000+5;
int H[MAXN];
int main(){
    int n, h;
    while (~scanf("%d%d", &n, &h)){
        int ans = 0;
        for (int i = 0; i < n; i++){
            scanf("%d", &H[i]);
            if (H[i]>h){
                ans += 2;
            }
            else{
                ans += 1;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}