#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, a, b, diff = 0x3f3f3f3f;
  scanf("%d", &n);
  int v[n];
  for (int i = (int)(0); i < (int)(n); ++i) scanf("%d", &v[i]);
  for (int i = (int)(0); i < (int)(n); ++i) {
    if (i == n - 1) {
      if (diff > abs(v[i] - v[0])) {
        diff = abs(v[i] - v[0]);
        a = i + 1;
        b = 1;
      }
    } else {
      if (diff > abs(v[i] - v[i + 1])) {
        diff = abs(v[i] - v[i + 1]);
        a = i + 1;
        b = i + 2;
      }
    }
  }
  cout << a << " " << b << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, a, b, diff = 0x3f3f3f3f;
  scanf("%d", &n);
  int v[n];
  for (int i = (int)(0); i < (int)(n); ++i) scanf("%d", &v[i]);
  for (int i = (int)(0); i < (int)(n); ++i) {
    if (i == n - 1) {
      if (diff > abs(v[i] - v[0])) {
        diff = abs(v[i] - v[0]);
        a = i + 1;
        b = 1;
      }
    } else {
      if (diff > abs(v[i] - v[i + 1])) {
        diff = abs(v[i] - v[i + 1]);
        a = i + 1;
        b = i + 2;
      }
    }
  }
  cout << a << " " << b << endl;
  return 0;
}

```