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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author PMarecki
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int M = (int) 1e6 + 10;
            TaskB.Segment[] ss = new TaskB.Segment[M];
            for (int i = 0; i < M; i++) ss[i] = new TaskB.Segment();
            int n = in.nextInt();
            for (int i = 0; i < n; i++) {
                int a = in.nextInt();
                ss[a].L = Math.min(ss[a].L, i);
                ss[a].R = Math.max(ss[a].R, i);
                ss[a].cnt++;
            }
            Arrays.sort(ss, new Comparator<TaskB.Segment>() {

                public int compare(TaskB.Segment o1, TaskB.Segment o2) {
                    return Integer.compare(o1.cnt, o2.cnt);
                }
            });
            TaskB.Segment best = ss[M - 1];
            for (int i = M - 1; i >= 0; --i) {
                if (ss[i].cnt != best.cnt) break;
                if (ss[i].R - ss[i].L < best.R - best.L) best = ss[i];
            }
            out.printf("%d %d\n", best.L + 1, best.R + 1);

        }

        class Segment {
            int L = 2000000;
            int R = -1;
            int cnt = 0;

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}