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 int mod = 1e9 + 7;
long long power(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= (int)1e9 + 7) ret %= (int)1e9 + 7;
    if (a >= (int)1e9 + 7) a %= (int)1e9 + 7;
    b >>= 1;
  }
  return ret;
}
long long inv(long long x) { return power(x, (int)1e9 + 7 - 2); }
int main() {
  long long n, min = (int)1e9 + 7, k = 0;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    if (abs(a[i] - a[i + 1]) < min) {
      min = abs(a[i] - a[i + 1]);
      k = i;
    }
  }
  if (abs(a[0] - a[n - 1]) < min) {
    cout << 1 << " " << n << endl;
    return 0;
  }
  cout << k + 1 << " " << k + 2 << endl;
}
