#include <bits/stdc++.h>
using namespace std;
void printArray(long long arr[], long long size) {
  for (int i = 0; i < size; i++) cout << arr[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, d = 100005;
  cin >> n;
  long long b[n + 1];
  long long p1, p2;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  b[n] = b[0];
  for (int i = 1; i < n + 1; i++) {
    if (abs(b[i] - b[i - 1]) < d) {
      d = abs(b[i] - b[i - 1]);
      p1 = i;
      p2 = i + 1;
    }
  }
  if (p1 == n && p2 == n + 1) p2 -= n;
  std::cout << p1 << " " << p2 << std::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;
void printArray(long long arr[], long long size) {
  for (int i = 0; i < size; i++) cout << arr[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, d = 100005;
  cin >> n;
  long long b[n + 1];
  long long p1, p2;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  b[n] = b[0];
  for (int i = 1; i < n + 1; i++) {
    if (abs(b[i] - b[i - 1]) < d) {
      d = abs(b[i] - b[i - 1]);
      p1 = i;
      p2 = i + 1;
    }
  }
  if (p1 == n && p2 == n + 1) p2 -= n;
  std::cout << p1 << " " << p2 << std::endl;
  return 0;
}

```