Little girl Susie went shopping with her mom and she wondered how to improve service quality.

There are n people in the queue. For each person we know time t_i needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.

Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.

The first line contains integer n (1 ≤ n ≤ 10⁵).

The next line contains n integers t_i (1 ≤ t_i ≤ 10⁹), separated by spaces.

Print a single number — the maximum number of not disappointed people in the queue.

5
15 2 1 5 3

4

Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.

題目大意：

一共有N個人要排隊，對應每個人如果等待的時間超過了自己的忍耐時間，那麼對應這個人就會走，問怎樣分配這些人進行排隊能夠使得最多的人得到服務。

思路（這個D好JB水啊）：

直接將每個人的忍耐時間從小到大排序。

接下來我們模擬整個排隊過程即可。

Ac程式碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll __int64
ll a[100060];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        int tmp=n;
        ll sum=0;
        sort(a,a+n);
        for(int i=0;i<n;i++)
        {
            if(sum<=a[i])
            {
                sum+=a[i];
            }
            else
            {
                tmp--;
            }
        }
        printf("%d\n",tmp);
    }
}