B.Watering System

B. Watering System

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.

Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1,s_2,\dots ,s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i\cdot A}{S}$ liters of water will flow out of it.

What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?

Input

The first line contains three integers $n$, $A$, $B$ ($1\le n\le 100000$, $1\le B\le A\le {10}^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.

The second line contains $n$ integers $s_1,s_2,\dots ,s_n$ ($1\le s_i\le {10}^4$) — the sizes of the holes.

Output

Print a single integer — the number of holes Arkady should block.

Examples

input

Copy

4 10 3
2 2 2 2

output

Copy

1

input

Copy

4 80 20
3 2 1 4

output

Copy

0

input

Copy

5 10 10
1000 1 1 1 1

output

Copy

4

Note

In the first example Arkady should block at least one hole. After that, $\frac{10\cdot 2}{6}\approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.

In the second example even without blocking any hole, $\frac{80\cdot 3}{10}=24$ liters will flow out of the first hole, that is not less than $20$.

In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.

题意：一共有n个洞，每个洞有一个s值，记没有被堵上的洞总的s值为sum，问你至少填几个洞能使a*S0/sum>=b

题解：直接排序贪心即可（交的时候数组开小了一直超时QWQ哎）

#include<iostream>  
#include<string.h>  
#include<algorithm>  
#include<cmath>  
#include<map>  
#include<string>  
#include<stdio.h>  
#include<vector>  
#include<stack>
#include<set>
using namespace std;
#define INIT ios::sync_with_stdio(false)
#define LL long long int
struct node {
	int h, m;
};
int n, a, b;
int num[1000005];
bool cmp(int s, int t) {
	return s > t;
}

int main() {


	while (cin >> n >> a >> b) {
		int sum = 0;
		for (int i = 0;i < n;i++) {
			scanf("%d", &num[i]);
			sum += num[i];
		}
		if (((a * num[0]) / sum) >= b) {
			cout << 0 << endl;
			continue;
		}
		sort(num + 1, num + n, cmp);
		int years = 0;
		for (int i = 1;i < n;i++) {
			if (((a * num[0]) / sum) >= b) {
				break;
			}
			sum = num[i];
			years++;
		}
		cout << ans << endl;
	}
	return 0;
}