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 = (int)1e6 + 123;
const int MOD = (int)1e9 + 7;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
void MaxMercury() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > a(n);
  pair<int, int> ans;
  int mn = 1001;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i + 1;
  }
  for (int i = 1; i < n; i++) {
    if (mn > abs(a[i].first - a[i - 1].first)) {
      mn = abs(a[i].first - a[i - 1].first);
      ans = make_pair(a[i].second, a[i - 1].second);
    }
  }
  if (mn > abs(a[0].first - a[n - 1].first)) {
    mn = abs(a[0].first - a[n - 1].first);
    ans = make_pair(a[0].second, a[n - 1].second);
  }
  cout << ans.first << " " << ans.second << endl;
}
int main() {
  MaxMercury();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
