#228(div2)C. Fox and Box Accumulation

C. Fox and Box Accumulation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most x_i boxes on its top (we'll call x_i the strength of the box).

Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.

Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than x_i boxes on the top of i-th box. What is the minimal number of piles she needs to construct?

Input

The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x₁, x₂, ..., x_n (0 ≤ x_i ≤ 100).

Output

Output a single integer — the minimal possible number of piles.

Examples

input

3
0 0 10

output

2

input

5
0 1 2 3 4

output

1

input

4
0 0 0 0

output

4

input

9
0 1 0 2 0 1 1 2 10

output

3

题意：给出每个相同盒子的强度（强度代表上面能累积几个盒子），问有几组
思路：我们知道最正常的就是012345.....那么我们可以得到当前第一个大于=当前i的盒子，贪心

#include<bits/stdc++.h>
using namespace std;

int a[103];
int sum=0;
int n;

void hh(){
    sort(a+1,a+1+n);
    for(int i=0;i<n;i++){
        int k=lower_bound(a+1,a+1+n,i)-a;
        if(k!=n+1){
            sum++;
            a[k]=-1;
        }
    }
}

int main(){

    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    int s=0;
    sort(a+1,a+1+n);
    while(1){
        s++;
        hh();
        if(sum==n) break;
    }
    cout<<s<<endl;
}