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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
inline void arin(T* st, T* nd) {
  while (st < nd) cin >> *st++;
}
int di[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dj[] = {0, 0, 1, -1, 1, -1, -1, 1};
int diK[] = {-2, -2, -1, 1, 2, 2, 1, -1};
int djK[] = {-1, 1, 2, 2, 1, -1, -2, -2};
bool cmd(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second < b.first.second);
}
int main() {
  int n;
  cin >> n;
  int a[n];
  arin(a, a + n);
  int l = 0, r = -1, def = 1e6;
  for (int j = 0; j < n; j++) {
    if (abs(a[j % n] - a[(j + 1) % n]) < def)
      l = j, r = (j + 1) % n, def = abs(a[j] - a[(j + 1) % n]);
  }
  cout << l + 1 << " " << r + 1;
}
