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>
void task_a() {
  int t = 0;
  std::cin >> t;
  for (int test = 0; test < t; ++test) {
    int num = 0;
    std::cin >> num;
    int high = 0;
    while ((1 << high) <= num) {
      high += 1;
    }
    int64_t bin_sum = (static_cast<int64_t>(1) << (high + 1)) - 2;
    int64_t result = (int64_t)num * (num + 1) / 2 - bin_sum;
    std::cout << result << std::endl;
  }
}
void task_b() {
  std::string s;
  std::cin >> s;
  std::string aux = s;
  int queries = 0;
  std::cin >> queries;
  for (int query = 0; query < queries; ++query) {
    int left = 0, right = 0, shift = 0;
    std::cin >> left >> right >> shift;
    left -= 1;
    right -= 1;
    for (int it = left; it <= right; ++it) {
      aux[left + (it - left + shift) % (right - left + 1)] = s[it];
    }
    s = aux;
  }
  std::cout << s << std::endl;
}
struct OrientedSegment {
  OrientedSegment(int _x = 1, int _y = 1, int _id = 0)
      : x(_x), y(_y), id(_id), phi(atan2(_y, _x)) {}
  bool operator<(const OrientedSegment &other) const { return phi < other.phi; }
  int x, y, id;
  long double phi;
};
void task_c() {
  int n = 0;
  std::cin >> n;
  std::vector<OrientedSegment> segments;
  segments.reserve(n);
  for (int i = 0; i < n; ++i) {
    int x = 0, y = 0;
    std::cin >> x >> y;
    segments.push_back(OrientedSegment(x, y, i + 1));
  }
  std::sort(segments.begin(), segments.end());
  int ans1 = -1, ans2 = -1;
  long double min = std::numeric_limits<long double>::max();
  for (int i = 0; i < n - 1; ++i) {
    if (segments[(i + 1) % n].phi - segments[i].phi < min) {
      min = segments[(i + 1) % n].phi - segments[i].phi;
      ans1 = segments[i].id;
      ans2 = segments[(i + 1) % n].id;
    }
  }
  if (segments[0].phi + 2 * 3.14159265358979323846 - segments[n - 1].phi <
      min) {
    min = segments[0].phi + 2 * 3.14159265358979323846 - segments[n - 1].phi;
    ans1 = segments[n - 1].id;
    ans2 = segments[0].id;
  }
  std::cout << ans1 << " " << ans2 << std::endl;
}
struct Cell {
  Cell(int _x, int _y) : x(_x), y(_y) {}
  int x, y;
};
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
bool is_in_field(int i, int j, int n, int m) {
  return 0 <= i && i < n && 0 <= j && j < m;
}
void bfs(int i, int j, int n, int m, std::vector<std::vector<int>> &result,
         std::vector<std::vector<bool>> &used,
         const std::vector<std::vector<char>> &field) {
  std::queue<Cell> queue;
  used[i][j] = true;
  queue.push(Cell(i, j));
  std::vector<Cell> path;
  path.push_back(Cell(i, j));
  int ans = 0;
  while (!queue.empty()) {
    Cell current = queue.front();
    queue.pop();
    for (int dir = 0; dir < 4; ++dir) {
      if (!is_in_field(current.x + dx[dir], current.y + dy[dir], n, m)) {
        continue;
      }
      if (field[current.x + dx[dir]][current.y + dy[dir]] == '*') {
        ans += 1;
      } else if (!used[current.x + dx[dir]][current.y + dy[dir]]) {
        used[current.x + dx[dir]][current.y + dy[dir]] = true;
        queue.push(Cell(current.x + dx[dir], current.y + dy[dir]));
        path.push_back(Cell(current.x + dx[dir], current.y + dy[dir]));
      }
    }
  }
  for (const auto &step : path) {
    result[step.x][step.y] = ans;
  }
}
std::vector<std::vector<int>> run_bfs(
    int n, int m, const std::vector<std::vector<char>> &field) {
  std::vector<std::vector<int>> result(n, std::vector<int>(m));
  std::vector<std::vector<bool>> used(n, std::vector<bool>(m, false));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (!used[i][j] && field[i][j] != '*') {
        bfs(i, j, n, m, result, used, field);
      }
    }
  }
  return result;
}
void task_d() {
  int n = 0, m = 0, k = 0;
  std::cin >> n >> m >> k;
  std::vector<std::vector<char>> field(n, std::vector<char>(m));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      std::cin >> field[i][j];
    }
  }
  std::vector<std::vector<int>> ans = run_bfs(n, m, field);
  for (int query = 0; query < k; ++query) {
    int x = 0, y = 0;
    std::cin >> x >> y;
    x -= 1;
    y -= 1;
    std::cout << ans[x][y] << std::endl;
  }
}
int main() {
  task_c();
  return 0;
}