Queue （周赛3）

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 timet_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.

Input

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

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

Output

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

Sample test(s)

Input

515 2 1 5 3

Output

4

Note

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.

为什么用Ｃ的快排就是超时，用Ｃ++的啥事儿没有，歧视有木有~      表示很醉~      = =！

#include <stdio.h>#include <iostream>#include <algorithm>using namespace std;int a[100000];int main(){    int n,i;    while(~scanf("%d",&n))    {        long long count=0,sum=0;        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        sort(a,a+n);  //快排        for(i=0;i<n;i++)        {            if(a[i]>=sum)            {                 count++;  //累加                sum=sum+a[i];  //之前的时间累加            }        }        printf("%lld\n",count);    }    return 0;}