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;
const int MOD = 1e6 + 3;
const int INFI = 1e9 * 2;
const int N = 111;
const int M = 222;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};
int a[N];
int abs(int x) { return x > 0 ? x : -x; }
int main() {
  int n, ans, x, y;
  while (scanf("%d", &n) != EOF) {
    ans = INFI;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < n; i++)
      if (ans > abs(a[i + 1] - a[i])) {
        ans = abs(a[i + 1] - a[i]);
        x = i;
        y = i + 1;
      }
    if (abs(a[n] - a[1]) < ans) x = n, y = 1;
    printf("%d ", x), printf("%d\n", y);
  }
  return 0;
}
