n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.

Output

Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.

Examples

Input

5
10 12 13 15 10


Output

5 1


Input

4
10 20 30 40


Output

1 2
#include <bits/stdc++.h>
using namespace std;
class AReconnaissance2 {
 public:
  void solve(std::istream &in, std::ostream &out) {
    int n;
    int arr[101];
    in >> n;
    for (int i = 1; i <= n; i++) in >> arr[i];
    int mn = 1000000000;
    int ind1, ind2;
    for (int i = 1; i < n; i++) {
      if (abs(arr[i] - arr[i + 1]) < mn) {
        mn = abs(arr[i] - arr[i + 1]);
        ind1 = i;
        ind2 = i + 1;
      }
    }
    if (abs(arr[1] - arr[n]) < mn) {
      ind1 = 1;
      ind2 = n;
    }
    out << ind1 << " " << ind2 << endl;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  AReconnaissance2 solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  in.tie(0);
  solver.solve(in, out);
  return 0;
}
