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>
using namespace std;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int mod = (int)1e9 + 7;
const long long INF64 = 3e18;
void smxl(long long &a, long long b) {
  if (a < b) a = b;
}
void smnl(long long &a, long long b) {
  if (a > b) a = b;
}
void adsl(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void misl(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
void smx(long long &a, long long b) {
  if (a < b) a = b;
}
void smn(long long &a, long long b) {
  if (a > b) a = b;
}
void ads(long long &a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
void mis(long long &a, long long b) {
  a -= b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long egcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long mbinp(long long a, long long b) {
  a %= mod;
  if (b == 0) return 1;
  long long ans = mbinp(a, b / 2);
  long long tmp = (ans * ans) % mod;
  if (b % 2) return ((tmp * a) % mod);
  return ((tmp) % mod);
}
long long binp(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = binp(a, b / 2);
  long long tmp = (ans * ans);
  if (b % 2) return ((tmp * a));
  return ((tmp));
}
long long C(int n, int m) {
  long long ret = 1;
  for (int i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long overbinp(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      if (res < INF64 / a)
        res *= a;
      else
        return INF64;
    }
    if (b > 1) {
      if (a < INF64 / a)
        a *= a;
      else
        return INF64;
    }
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  vector<pair<long double, int> > deg;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    deg.push_back(make_pair(
        (atan2(b, a) < 0 ? atan2(b, a) + 2 * acos(-1) : atan2(b, a)), i));
  }
  sort(deg.begin(), deg.end());
  long double mx = deg[n - 1].first - deg[0].first;
  if (mx > acos(-1)) mx = 2 * acos(-1) - mx;
  int res1 = deg[0].second;
  int res2 = deg[n - 1].second;
  for (int i = 0; i < n - 1; i++) {
    long double tmp = deg[i + 1].first - deg[i].first;
    if (tmp > acos(-1)) tmp = 2 * acos(-1) - tmp;
    if (tmp < mx) {
      mx = tmp;
      res1 = deg[i + 1].second;
      res2 = deg[i].second;
    }
  }
  cout << res1 + 1 << " " << res2 + 1 << endl;
}