#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, minn = 0x3f3f3f3f;
  int a[110];
  int b[110] = {0};
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int tmpx, tmpy;
  for (int i = 1; i <= n; i++) {
    if (i == n) {
      b[i] = abs(a[i] - a[1]);
    } else {
      b[i] = abs(a[i] - a[i + 1]);
    }
    if (b[i] <= minn) {
      minn = b[i];
      if (i == n) {
        tmpx = n;
        tmpy = 1;
      } else {
        tmpx = i;
        tmpy = i + 1;
      }
    }
  }
  printf("%d %d\n", tmpx, tmpy);
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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() {
  int n, minn = 0x3f3f3f3f;
  int a[110];
  int b[110] = {0};
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int tmpx, tmpy;
  for (int i = 1; i <= n; i++) {
    if (i == n) {
      b[i] = abs(a[i] - a[1]);
    } else {
      b[i] = abs(a[i] - a[i + 1]);
    }
    if (b[i] <= minn) {
      minn = b[i];
      if (i == n) {
        tmpx = n;
        tmpy = 1;
      } else {
        tmpx = i;
        tmpy = i + 1;
      }
    }
  }
  printf("%d %d\n", tmpx, tmpy);
  return 0;
}

```