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;
  int array[110];
  int sum[110];
  int a, b;
  int i, j, k;
  int temp, min;
  while (scanf("%d", &n) != EOF) {
    for (i = 1; i <= n; i++) {
      scanf("%d", &array[i]);
    }
    for (i = 1; i <= n; i++) {
      if (i == 1) {
        min = array[i + 1] - array[i];
        if (min < 0) min = (-1) * min;
        a = 1;
        b = 2;
      } else if (i != n) {
        temp = array[i + 1] - array[i];
        if (temp < 0) temp = (-1) * temp;
        if (temp < min) {
          min = temp;
          a = i;
          b = i + 1;
        }
      } else if (i == n) {
        temp = array[n] - array[1];
        if (temp < 0) temp = (-1) * temp;
        if (temp < min) {
          min = temp;
          a = n;
          b = 1;
        }
      }
    }
    printf("%d %d\n", a, b);
  }
  return 0;
}
