Watering System （模拟，水题）

Watering System

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 s1,s2,…,sn. In other words, if the sum of sizes of non-blocked holes is S, and the i-th hole is not blocked, si⋅AS

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≤n≤100000, 1≤B≤A≤104

) — 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 s1,s2,…,sn ( 1≤si≤104

) — the sizes of the holes.

Output

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

Examples

Input

4 10 3
2 2 2 2

Output

1

Input

4 80 20
3 2 1 4

Output

0

Input

5 10 10
1000 1 1 1 1

Output

4

Note

In the first example Arkady should block at least one hole. After that, 10⋅26≈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, 80⋅310=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.

思路：直接从第二个到最后一个进行从小到大排序，然后依次删除s[i]，计算s1×A/sum，判断是否大于等于B

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5+10;
bool cmp(int a,int b){
    return a > b;
}
int main(){
    int n,a,b;
    int s[maxn];
    int sum = 0;
    cin >> n >> a >> b;
    for(int i = 1; i <= n; i++){
        cin >> s[i];
        sum += s[i];
    }
    if(1.0 * s[1] * a / sum >= b){
        cout << "0" << endl;
        return 0;
    }
    sort(s+2,s+1+n,cmp);
    int cnt = 0;
    for(int i = 2; i <= n; i++){
        sum -= s[i];
        cnt++;
        if(1.0 * s[1] * a / sum >= b){
            cout << cnt << endl;
            break;
        }
    }
    return 0;
}