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)

// TOPSIC001_0

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// mod
const ll MOD = 1000000007ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

// floating
typedef double Real;
const Real EPS = 1e-11;
#define EQ0(x) (abs(x)<EPS)
#define EQ(a,b) (abs(a-b)<EPS)
typedef complex<Real> P;


int main(){
    double N,T,A; cin >> N >> T >> A;
//    T *= 1000;

    vi H(N); REP(i, N) cin >> H.at(i);

    double ans,temp = 10000;
    REP(i,N){
        long tm = abs((T - H.at(i) * 0.006) - A);
        if(tm  < temp){
            ans = i;
            temp = tm;
        }
    }

    cout << ans + 1;

    return 0;
}