#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n;
  cin >> n;
  long long int i, mn = INT_MAX, a, b;
  long long int arr[n];
  for (i = 0; i <= n - 1; i++) {
    cin >> arr[i];
  }
  for (i = 0; i <= n - 1; i++) {
    if (i == 0) {
      if (abs(arr[n - 1] - arr[0]) < mn) {
        mn = abs(arr[n - 1] - arr[0]);
        a = n;
        b = 1;
      }
    }
    if (i >= 1) {
      if (abs(arr[i] - arr[i - 1]) < mn) {
        mn = abs(arr[i] - arr[i - 1]);
        a = i;
        b = i + 1;
      }
    }
  }
  cout << a << " " << b << "\n";
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n;
  cin >> n;
  long long int i, mn = INT_MAX, a, b;
  long long int arr[n];
  for (i = 0; i <= n - 1; i++) {
    cin >> arr[i];
  }
  for (i = 0; i <= n - 1; i++) {
    if (i == 0) {
      if (abs(arr[n - 1] - arr[0]) < mn) {
        mn = abs(arr[n - 1] - arr[0]);
        a = n;
        b = 1;
      }
    }
    if (i >= 1) {
      if (abs(arr[i] - arr[i - 1]) < mn) {
        mn = abs(arr[i] - arr[i - 1]);
        a = i;
        b = i + 1;
      }
    }
  }
  cout << a << " " << b << "\n";
  return 0;
}

```