Problem: Amr has got a large array of size n. Amr doesn't like large arrays so he intends to make it smaller.

Amr doesn't care about anything in the array except the beauty of it. The beauty of the array is defined to be the maximum number of times that some number occurs in this array. He wants to choose the smallest subsegment of this array such that the beauty of it will be the same as the original array.

Help Amr by choosing the smallest subsegment possible.

Input

The first line contains one number n (1 ≤ n ≤ 105), the size of the array.

The second line contains n integers ai (1 ≤ ai ≤ 106), representing elements of the array.

Output

Output two integers l, r (1 ≤ l ≤ r ≤ n), the beginning and the end of the subsegment chosen respectively.

If there are several possible answers you may output any of them. 

Examples

Input

5
1 1 2 2 1


Output

1 5

Input

5
1 2 2 3 1


Output

2 3

Input

6
1 2 2 1 1 2


Output

1 5

Note

A subsegment B of an array A from l to r is an array of size r - l + 1 where Bi = Al + i - 1 for all 1 ≤ i ≤ r - l + 1

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class AmrandTheLargeArray {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok;

  static void solve() throws Exception {
    int n = nextInt();
    int[] a = nextIntArray(n + 1, 1);
    Map<Integer, LeftRightCount> map = new HashMap<>();
    for (int i = 1; i <= n; i++) {
      if (map.containsKey(a[i])) {
        LeftRightCount lrc = map.get(a[i]);
        lrc.right = i;
        lrc.count++;
      } else {
        LeftRightCount lrc = new LeftRightCount(i, i, 1);
        map.put(a[i], lrc);
      }
    }
    LeftRightCount ans = new LeftRightCount(0, 0, 0);
    for (LeftRightCount lrc : map.values()) {
      if (lrc.count > ans.count)
        ans = lrc;
      else if (lrc.count == ans.count && ((ans.right - ans.left)) > (lrc.right - lrc.left))
        ans = lrc;
    }
    out.println(ans.left + " " + ans.right);
  }

  public static void main(String args[]) {
    try {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(new OutputStreamWriter(System.out));
      solve();
      in.close();
      out.close();
    } catch (Throwable e) {
      e.printStackTrace();
      System.exit(1);
    }
  }

  static int nextInt() throws IOException {
    return Integer.parseInt(next());
  }

  static int[] nextIntArray(int len, int start) throws IOException {
    int[] a = new int[len];
    for (int i = start; i < len; i++)
      a[i] = nextInt();
    return a;
  }

  static long nextLong() throws IOException {
    return Long.parseLong(next());
  }

  static long[] nextLongArray(int len, int start) throws IOException {
    long[] a = new long[len];
    for (int i = start; i < len; i++)
      a[i] = nextLong();
    return a;
  }

  static String next() throws IOException {
    while (tok == null || !tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

}


class LeftRightCount {
  int left;
  int right;
  int count;

  LeftRightCount(int a, int b, int c) {
    left = a;
    right = b;
    count = c;
  }
}