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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
/*
    // 配列の最大値を取得
    int max = *max_element(vec.begin(), vec.end());
    // 配列の最大値のイテレータを取得
    vector<int>::iterator maxIt = max_element(vec.begin(), vec.end());
    // 配列の最大値までの距離を取得
    int maxIndex = distance(vec.begin(), maxIt);
*/

int main () {
    int N; cin >> N;
    int T, A; cin >> T >> A;
    vector<int> H(N);
    for (int i=0; i<N; ++i) cin >> H[i];

    vector<int> temp;
    for (int i=0; i<N; ++i) {
        int t = T - H[i] * 0.006;
        temp.push_back(abs(t - A));
    }

    vector<int>::iterator minIt = min_element(temp.begin(), temp.end());
    cout << distance(temp.begin(), minIt) + 1 << endl;
}