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;
int bs(vector<long long>& a, int que) {
  auto it = lower_bound(a.begin(), a.end(), que);
  if (*it == que) {
    return it - a.begin();
  }
  return -1;
}
void subsets(int S[], int n) {
  int nSub = 1 << n;
  for (int i = 0; i < nSub; i++) {
    cout << "{ ";
    for (int k = 0; k < n; k++) {
      if ((1 << k) & i) {
        cout << S[k] << " ";
      }
    }
    cout << "}\n";
  }
}
void permutate() {
  int n, x;
  cin >> n;
  vector<int> permutation;
  for (int i = 0; i < n; i++) {
    cin >> x;
    permutation.push_back(x);
  }
  do {
    for (int i = 0; i <= n - 1; i++) cout << permutation[i];
    cout << endl;
  } while (next_permutation(permutation.begin(), permutation.end()));
}
bool prime(long long x) {
  long long ans = 0;
  if (x == 1) {
    return false;
  }
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  int n, min, ind1, ind2;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  min = abs(a[0] - a[n - 1]);
  ind1 = 1;
  ind2 = n;
  for (int i = 0; i < n - 1; i++) {
    if (min > abs(a[i] - a[i + 1])) {
      min = abs(a[i] - a[i + 1]);
      ind1 = i + 1;
      ind2 = i + 2;
    }
  }
  cout << ind1 << " " << ind2;
  return 0;
}
