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 long long N = 1e6 + 10;
const long long INF = 1e18;
const long long inf = -1e18;
const long long MOD = 1e9 + 7;
const long long base = 1000 * 1000 * 1000;
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  long long mini = 1e9 + 7;
  long long s, e;
  for (long long i = 0; i < n - 1; i++) {
    if (mini > abs(a[i] - a[i + 1])) {
      mini = abs(a[i] - a[i + 1]);
      s = i + 1;
      e = i + 2;
    }
  }
  if (mini > abs(a[0] - a[n - 1])) {
    s = 1;
    e = n;
  }
  cout << s << " " << e << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}
