Problem: You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.

Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 and π. For example, opposite directions vectors have angle equals to π.

Input

First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.

The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed that no two vectors in the input share the same direction (but they still can have opposite directions).

Output

Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any.

Examples

Input

4
-1 0
0 -1
1 0
1 1


Output

3 4


Input

6
-1 0
0 -1
1 0
1 1
-4 -5
-4 -6


Output

6 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
using dou = long double;
string yes = "yes";
string Yes = "Yes";
string YES = "YES";
string no = "no";
string No = "No";
string NO = "NO";
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long mod = 1000000007ll;
struct mint {
  long long x;
  mint(long long x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(long long t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
long long ceili(long long a, long long b) { return ((a + b - 1) / b); }
const int INF = 2000000000;
const long long INF64 = 4223372036854775807ll;
const long long MOD = 1000000007ll;
const long long OD = 1000000000000007ll;
const dou pi = 3.141592653589793;
long long modpow(long long a, long long n) {
  long long resize = 1;
  while (n > 0) {
    if (n & 1) resize = resize * a % MOD;
    a = a * a % MOD;
    n >>= 1;
  }
  return resize;
}
int main() {
  int n;
  std::cin >> n;
  std::vector<dou> x(n), y(n);
  for (long long i = 0; i < (long long)(n); i++) std::cin >> x[i] >> y[i];
  std::vector<pair<dou, int>> p(n);
  for (long long i = 0; i < (long long)(n); i++) {
    p[i].second = i + 1;
    p[i].first = atan2(x[i], y[i]);
  }
  sort((p).begin(), (p).end());
  p.push_back(p[0]);
  dou ma = 123456;
  int an, ann;
  for (long long i = 0; i < (long long)(n); i++) {
    if (chmin(ma, min(2 * acos(-1) - abs(p[i].first - p[i + 1].first),
                      abs(p[i].first - p[i + 1].first)))) {
      an = p[i].second;
      ann = p[i + 1].second;
    }
  }
  std::cout << an << " " << ann << std::endl;
}