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;
const int MAXN = 2e5 + 5;
const unsigned int M = 1000000007;
priority_queue<int, vector<int>, greater<int> > q;
vector<int> seq;
vector<int> graph[100005];
bool visit[100005];
bool achie[201];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int a[201];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int mini = INT_MAX;
  int row, col;
  for (int i = 0; i < n + 1; i++) {
    if (abs(a[i % n] - a[(i + 1) % n]) < mini) {
      mini = abs(a[i % n] - a[(i + 1) % n]);
      row = i % n;
      col = (i + 1) % n;
    }
  }
  cout << row + 1 << " " << col + 1 << "\n";
}
