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;
void adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int MAX = 2e5 + 7;
const int MOD = 1e+9 + 7;
const long long INF = 1e+18 + 7;
int nod(int a, int b) {
  if (a == 0 || b == 0) {
    return 0;
  } else if (a == b) {
    return a;
  } else if (a > b) {
    return nod(a - b, b);
  } else {
    return nod(a, b - a);
  }
}
int nok(int a, int b) { return a * b / nod(a, b); }
void sp(int a, double b) { cout << fixed << setprecision(a) << b; }
int a[1000000];
void solve() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
  }
  int ans = 10000000;
  for (int i = (0); i < (n - 1); i++) {
    if (abs(a[i] - a[i + 1]) == 0) {
      cout << i + 1 << " " << i + 2;
      return;
    } else {
      ans = min(ans, abs(a[i] - a[i + 1]));
    }
  }
  ans = min(ans, abs(a[0] - a[n - 1]));
  if (abs(a[0] - a[n - 1]) == 0 || ans == abs(a[0] - a[n - 1])) {
    cout << 1 << " " << n;
    return;
  }
  for (int i = (0); i < (n - 1); i++) {
    if (abs(a[i] - a[i + 1]) == ans) {
      cout << i + 1 << " " << i + 2;
      return;
    }
  }
}
int main() {
  adskiy_razgon();
  long long t = 1;
  for (int i = (1); i < (t + 1); i++) {
    solve();
  }
  return 0;
}
