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 comment(linker, "/stack:640000000")
using namespace std;
const double EPS = 1e-9;
const int INF = 0x7f7f7f7f;
const double PI = acos(-1.0);
template <class T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int t, n, m;
int arr[105];
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < (n); i++) scanf("%d", &arr[i]);
    pair<int, int> ans;
    int cur = INF;
    for (int i = 0; i < (n); i++) {
      if (abs(arr[i] - arr[(i + 1) % n]) < cur) {
        cur = abs(arr[i] - arr[(i + 1) % n]);
        ans = make_pair(i, (i + 1) % n);
      }
    }
    ans.first++;
    ans.second++;
    printf("%d %d\n", ans.first, ans.second);
  }
  return 0;
}
