Problem:
Score : 400 points

Problem Statement

Snuke found N strange creatures.
Each creature has a fixed color and size. The color and size of the i-th creature are represented by i and A_i, respectively.

Every creature can absorb another creature whose size is at most twice the size of itself.
When a creature of size A and color B absorbs another creature of size C and color D (C \leq 2 \times A), they will merge into one creature of size A+C and color B.
Here, depending on the sizes of two creatures, it is possible that both of them can absorb the other.

Snuke has been watching these creatures merge over and over and ultimately become one creature.
Find the number of the possible colors of this creature.

Constraints

2 \leq N \leq 100000

1 \leq A_i \leq 10^9

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

Print the number of the possible colors of the last remaining creature after the N creatures repeatedly merge and ultimately become one creature.

Sample Input 1

3
3 1 4

Sample Output 1

2

The possible colors of the last remaining creature are colors 1 and 3.
For example, when the creature of color 3 absorbs the creature of color 2, then the creature of color 1 absorbs the creature of color 3, the color of the last remaining creature will be color 1.

Sample Input 2

5
1 1 1 1 1

Sample Output 2

5

There may be multiple creatures of the same size.

Sample Input 3

6
40 1 30 2 7 20

Sample Output 3

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<iomanip>
#include<limits>
#include<deque>
#include<map>
#include<list>
#include<set>
#include <unordered_set>
#include<vector>
#include<cmath>
#include<cstdio>
#include<memory>
#include<bitset>
#include<stack>
#include<functional>
#include<queue>
#include<regex>
#include<time.h>

using namespace std;


#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl

#define Tof(x) (x) ? Yes : No
#define TOF(x) (x) ? YES : NO
#define tof(x) (x) ? yes : no

using ll = long long;

//constexpr ll MOD = 1000000007;
constexpr ll INF = 1050000000;


/*vectorの要素を全部確認するやつだよ。普通に拡張for文したほうがいいよ。

for (vector<int>::iterator itr = ans.begin(); itr != ans.end(); ++itr) {
	cout << *itr << " ";
}

*/


/*for文でx,yを上下左右確認するやつだよ。正直使いづらいよ。

int dy[] = { 0, 1, 0, -1 };
int dx[] = { 1, 0, -1, 0 };


for (int i = 0; i < 4; i++) {
	int ny = y + dy[i];
	int nx = x + dx[i];
}

*/


/*繰り返し二乗法だよ。modもとってくれるよ。n^kをmodで割った余りでやってくれるよ。

ll POW_MOD(ll n, ll k, ll mod) {
	ll r = 1;

	for (; k > 0; k >>= 1) {
		if (k & 1) {
			r = (r * n) % mod;
		}
		n = (n * n) % mod;
	}
	return r;
}

*/


/*実行開始からの経過時間を表すよ。

while (clock() < 1900000) {  }

*/


/*まだ使えないテクニックのメモだよ。

for (int bit = 0; bit < (1<<n); ++bit)//bit全探索用のfor文だよ

*/


/*2重for文書くのめんどくなった時用だよ。

for (int i = 0; i < h; ++i) {
	for (int j = 0; j < w; ++j) {

	}
}

*/


int main() {
	int n; cin >> n;
	ll a[100005];

	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}

	sort(a, a + n);

	ll tmp = 0;

	for (int i = 0; i < n - 1; i++) {
		tmp += a[i];
		if (2 * tmp < a[i + 1]) { cout << n - (i + 1) << endl; return 0; }
	}

	cout << n << endl;


	return 0;
}