output_description: Print the index of the place where the palace should be built.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      T A
      H_1 H_2 ... H_N
attempt: #include "bits/stdc++.h"
  using namespace std;
  using ll = long long;
  using Graph = vector<vector<int>>;
  #define int ll
  #define double long double
  #define YES {cout<<"YES"<<'\n';}
  #define NO {cout<<"NO"<<'\n';}
  #define Yes {cout<<"Yes"<<'\n';}
  #define No {cout<<"No"<<'\n';}
  #define yes {cout<<"yes"<<'\n';}
  #define no {cout<<"no"<<'\n';}
  void YN(bool flg){cout<<(flg?"YES":"NO")<<'\n';}
  void Yn(bool flg){cout<<(flg?"Yes":"No")<<'\n';}
  void yn(bool flg){cout<<(flg?"yes":"no")<<'\n';}
  #define SORT(a) sort(a.begin(),a.end())
  #define REVERSE(a) reverse(a.begin(),a.end())
  #define REP(i, n) for(int i = 0; i < n; i++)
  #define REPR(i, n) for(int i = n; i >= 0; i--)
  #define FOR(i, m, n) for(int i = m; i < n; i++)
  #define out(n) {cout << n << '\n';}
  typedef vector<int> VI;
  typedef vector<string> VS;
  typedef vector<bool> VB;
  typedef vector<VI> VVI;
  const int MOD = 1000000007;
  const long long INF = 10e10;
  const double PI = acos(-1.0);
  template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
  template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
  template<class T>auto MAX(const T& a) { return *max_element(a.begin(),a.end()); }
  template<class T>auto MIN(const T& a) { return *min_element(a.begin(),a.end()); }
  template<class T, class U>U SUM(const T& a, const U& v) { return accumulate(a.begin(),a.end(), v); }
  template<class T, class U>U COUNT(const T& a, const U& v) { return count(a.begin(),a.end(), v); }
  template<class T, class U>int LOWER(const T& a, const U& v) { return lower_bound(a.begin(),a.end(), v) - a.begin(); }
  template<class T, class U>int UPPER(const T& a, const U& v) { return upper_bound(a.begin(),a.end(), v) - a.begin(); }
  int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }
  int LCM(int a, int b) { int g = GCD(a, b); return a / g * b; }
  int POW(int a, int n) { int r = 1; while (n > 0) { if (n & 1)r *= a; a *= a; n /= 2; }return r; }
  int isPrime(int n){if(n<2){return 0;}for(int i=2; i*i<=n; i++){if(n%i==0){return 0;}}return 1;}
  /*素因数分解*/map< int, int >prime_factor(int n){map< int, int > ret;for(int i = 2; i * i <= n; i++) {while(n % i == 0) {ret[i]++;n /= i;}}if(n != 1) ret[n] = 1;return ret;}
  /*約数列挙*/vector<int> divisor(int n){vector<int> v; for(int i=1; i*i<=n; i++){if(n%i==0){v.push_back(i);if(i!=n/i){v.push_back(n/i);}}}sort(v.begin(),v.end()); return v;}
  //----------------ライブラリとか----------------
  
  signed main(){
      cin.tie(0);
      ios::sync_with_stdio(0);
      cout<<fixed<<setprecision(15);
  
      int n,t,a; cin >> n >> t >> a;
      VI h(n); REP(i,n) cin >> h[i];
      //hをxに近づける
      int x = 6*(t-a)/1000;
      int ans , tmp = INF;
      REP(i,n){if(tmp>abs(h[i]-x)){tmp=abs(h[i]-x); ans=i+1;}}
      out(ans);
  
      return 0;
  }
problem_description: 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.
samples: [{"input": "2\n    12 5\n    1000 2000", "output": "1\n    \n\n  * The average temperature of Place 1 is 12-1000 \\times 0.006=6 degrees Celsius.\n  * The average temperature of Place 2 is 12-2000 \\times 0.006=0 degrees Celsius.\n\nThus, the palace should be built at Place 1.\n\n* * *"}, {"input": "3\n    21 -11\n    81234 94124 52141", "output": "3"}]