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;
using ll = long long int;
using ld = long double;
using vi = vector<int>;
using vll = vector<ll>;
using vd = vector<double>;
using vvi = vector<vi>;
int ctoi(char c) { return c - '0'; }
int slen(string s) { return s.length(); }
int vsize(vi a) { return (int)a.size(); }
int in() {
  int x;
  cin >> x;
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vi d(N);
  for (auto& x : d) cin >> x;
  int mn = abs(d[N - 1] - d[0]);
  int a = N - 1, b = 0;
  for (int i = 1; i < N; i++) {
    if (abs(d[i] - d[i - 1]) < mn) {
      mn = abs(d[i] - d[i - 1]);
      a = i;
      b = i - 1;
    }
  }
  cout << a + 1 << " " << b + 1 << endl;
  return 0;
}
