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;
using namespace std;
long long gcd(long long a, long long b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long pow(long long b, long long e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long a = pow(b, e / 2);
    return a * a;
  } else {
    long long a = pow(b, e / 2);
    return b * a * a;
  }
}
long long pow_m(long long x, long long y, long long m = 1000000007) {
  x = x % m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long modInverse(long long a, long long m) { return pow_m(a, m - 2, m); }
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long bin_coff(long long n, long long k, long long m) {
  vector<vector<long long>> ans(n + 1, vector<long long>(k, 0));
  for (long long i = 0; 0 < n + 1 ? i < n + 1 : i > n + 1;
       0 < n + 1 ? i += 1 : i -= 1) {
    for (long long j = 0;
         0 < min(i, k) + 1 ? j < min(i, k) + 1 : j > min(i, k) + 1;
         0 < min(i, k) + 1 ? j += 1 : j -= 1) {
      if (j == 0 || j == i)
        ans[i][j] = 1;
      else
        ans[i][j] = ans[i - 1][j - 1] % m + ans[i - 1][j] % m;
    }
  }
  return ans[n][k] % m;
}
long long inverse(long long i) {
  if (i == 1) return 1;
  return (1000000007 -
          ((1000000007 / i) * inverse(1000000007 % i)) % 1000000007 +
          1000000007) %
         1000000007;
}
bool prime(long long n) {
  if (n == 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0) return false;
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long tt = 1;
  for (long long i = 0; 0 < tt ? i < tt : i > tt; 0 < tt ? i += 1 : i -= 1) {
    long long n;
    cin >> n;
    vector<long long> arr(n);
    for (long long i = 0; i < n; i++) cin >> arr[i];
    long long diff = 1e18;
    long long k = 0, l = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (abs(arr[i + 1] - arr[i]) < diff) {
        k = i + 1;
        l = i;
        diff = abs(arr[i + 1] - arr[i]);
      }
    }
    if (abs(arr[0] - arr[n - 1]) < diff) {
      k = 0;
      l = n - 1;
    }
    cout << k + 1 << " " << l + 1 << endl;
  }
  return 0;
}
