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() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int m, i1, i2;
  if (arr[0] - arr[n - 1] > 0) {
    m = arr[0] - arr[n - 1];
    i1 = 1;
    i2 = n;
  } else if (arr[0] - arr[n - 1] < 0) {
    m = arr[n - 1] - arr[0];
    i1 = 1;
    i2 = n;
  }
  for (int j = 0; j < n - 1; j++) {
    int k = j + 1;
    if ((arr[j] - arr[k]) > 0) {
      if ((arr[j] - arr[k]) < m) {
        m = arr[j] - arr[k];
        i1 = j + 1;
        i2 = k + 1;
      }
    } else {
      if ((arr[k] - arr[j]) < m) {
        m = arr[k] - arr[j];
        i1 = j + 1;
        i2 = k + 1;
      }
    }
  }
  cout << i1 << " " << i2;
  return 0;
}
