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;
void fastscan(int &x) {
  bool neg = false;
  register int c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) x = (x << 1) + (x << 3) + c - 48;
  if (neg) x *= -1;
}
int main() {
  int n, arr[100];
  fastscan(n);
  for (int i = 0; i < n; i++) {
    fastscan(arr[i]);
  }
  int mindiff = INT_MAX;
  int key1, key2;
  for (int i = 0; i < n - 1; i++) {
    if (abs(arr[i] - arr[i + 1]) < mindiff) {
      mindiff = abs(arr[i] - arr[i + 1]);
      key1 = i + 1;
      key2 = i + 2;
    }
  }
  if (abs(arr[n - 1] - arr[0]) < mindiff) {
    key1 = 1;
    key2 = n;
  }
  cout << key1 << " " << key2;
  return 0;
}
