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;
int main() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long int> v;
  long long int mini = 99999999;
  for (long long int i = 0; i < (n - 1); i++) {
    v.push_back(abs(a[i + 1] - a[i]));
    mini = min(mini, v[i]);
  }
  v.push_back(abs(a[n - 1] - a[0]));
  mini = min(mini, v[v.size() - 1]);
  long long int ans;
  for (long long int i = 0; i < v.size(); i++) {
    if (v[i] == mini) {
      ans = i;
      break;
    }
  }
  if (ans <= (n - 2))
    cout << (ans + 1) << " " << (ans + 2) << "\n";
  else
    cout << (ans + 1) << " " << 1 << "\n";
}
