Problem:
Score: 200 points

Problem Statement

A country decides to build a palace.

In this country, the average temperature of a point at an elevation of x meters is T-x \times 0.006 degrees Celsius.

There are N places proposed for the place. The elevation of Place i is H_i meters.

Among them, Princess Joisino orders you to select the place whose average temperature is the closest to A degrees Celsius, and build the palace there.

Print the index of the place where the palace should be built.

It is guaranteed that the solution is unique.

Constraints

1 \leq N \leq 1000

0 \leq T \leq 50

-60 \leq A \leq T

0 \leq H_i \leq 10^5

All values in input are integers.

The solution is unique.

Input

Input is given from Standard Input in the following format:

N
T A
H_1 H_2 ... H_N

Output

Print the index of the place where the palace should be built.

Sample Input 1

2
12 5
1000 2000

Sample Output 1

1

The average temperature of Place 1 is 12-1000 \times 0.006=6 degrees Celsius.

The average temperature of Place 2 is 12-2000 \times 0.006=0 degrees Celsius.

Thus, the palace should be built at Place 1.

Sample Input 2

3
21 -11
81234 94124 52141

Sample Output 2

3

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

#include <iostream>
#include <string>
#include <cmath>
#include <vector>
#include <algorithm>
#include <iterator>
#include <typeinfo>
using namespace std;


// define list


namespace  math{
    const double PI = 3.14;
    const int INF = 1000000007;
    int GCM(int a,int b); //最大公約数
    int LCM(int a,int b); //最小公倍数
    int Pemu(int n,int r); //順列
    int Comb(int n,int m); //組み合わせ
    int Digit(int N); //桁和
    
    int GCM(int a,int b){
        if(a<b){
            swap(a,b);
        }
        int r = a%b;
        while (r!=0)
        {
            a=b;
            b=r;
            r=a%b;
        }
        return b;
    }

    int LCM(int a,int b){
        return a*(b/GCM(a,b));
    }

    int Pemu(int n,int r){
        int ans = n;
        for(int i=1;i<r;i++){
            ans*=(ans-i);
        }
        return ans;
    }

    int Comb(int n,int m){
        int a=n,b=m;
        for(int i=1;i<m;i++){
            a *= (a-i);
        }
        for(int i=1;i<m;i++){
            b *= (b-i);
        }
        
        return a/b;
    }

    int Digit(int N){
        if(N<10){
            return N;
        }
        int r = 0;
        while (N>=10)
        {
            r += (N%10);
            N /= 10;
        }
        r += N;
        return r;
    }
}

namespace SubLib{
    #define int long long
    #define double long double
    #define incin(n) int (n); cin >> (n);
    #define incin2(a,b) int a,b; cin >> a >> b; 
    #define rap(n,i) for(int (i) = 0;i<(n);(i)++)
    #define input(n,v) (v) = SubLib::Input((n),(v))
    #define all(v) (v).begin(),(v).end()
    using intv = vector<int>;
    using strv = vector<string>;
    using charv = vector<char>;
    template <class T1>
    using pvec = vector<pair<T1,T1>>;
    template <typename T> vector<T> Input(int n,vector<T> v){
        for(int i=0;i<n;i++){
            cin >> v.at(i);
        }
        return v;
    }
}

using namespace math;
using namespace SubLib;
double Kion(int X,int T){
    return T -X * 0.006;
}

signed main(){
    incin(n);
    incin2(t,a);
    intv h(n);
    input(n,h);
    int log_dif = 1000000000;
    int point;
    rap(n,i){   
        if(abs(Kion(h[i],t)-a)<log_dif){
            log_dif = abs(Kion(h[i],t)-a);
            point = i+1;
        }
    }
    cout << point << endl;
    return 0;
}