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>
int main() {
  int N;
  scanf("%d", &N);
  int a[N];
  scanf("%d", &a[0]);
  scanf("%d", &a[1]);
  int min = ((a[1] - a[0]) >= 0 ? (a[1] - a[0]) : (-1 * (a[1] - a[0])));
  int minindex = 0;
  for (int i = 2; i < N; i++) {
    scanf("%d", &a[i]);
    if (min > ((a[i] - a[i - 1]) >= 0 ? (a[i] - a[i - 1])
                                      : (-1 * (a[i] - a[i - 1])))) {
      min = ((a[i] - a[i - 1]) >= 0 ? (a[i] - a[i - 1])
                                    : (-1 * (a[i] - a[i - 1])));
      minindex = i - 1;
    }
  }
  if (min >
      ((a[0] - a[N - 1]) >= 0 ? (a[0] - a[N - 1]) : (-1 * (a[0] - a[N - 1])))) {
    min =
        ((a[0] - a[N - 1]) >= 0 ? (a[0] - a[N - 1]) : (-1 * (a[0] - a[N - 1])));
    minindex = N - 1;
  }
  printf("%d %d\n", minindex + 1, 1 + (minindex + 1) % N);
}
