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

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

class TaskB {

    static final int N = 1_000_001;

    public void solve(int testNumber, FastScanner in, PrintWriter out) {
        int n = in.nextInt();
        int[] a = in.nextIntArray(n);

        int[] count = new int[N];
        int[] start = new int[N];
        int[] end = new int[N];

        Arrays.fill(start, -1);
        Arrays.fill(end, -1);

        for (int i = 0; i < n; i++) {
            int num = a[i];
            count[num]++;
        }

        int max = -1;
        for (int i = 0; i < N; i++) {
            max = Math.max(max, count[i]);
        }

        for (int i = 0; i < n; i++) {
            int num = a[i];
            if (start[num] == -1)
                start[num] = i;
        }

        for (int i = n - 1; i >= 0; i--) {
            int num = a[i];
            if (end[num] == -1)
                end[num] = i;
        }

        int l = 1;
        int r = n;
        int length = n;
        for (int i = 0; i < N; i++) {
            if (count[i] == max) {
                int newLength = end[i] - start[i] + 1;
                if (newLength < length) {
                    l = start[i] + 1;
                    r = end[i] + 1;
                    length = newLength;
                }
            }
        }

        out.print(l + " " + r);
    }

}

class FastScanner {

    private final BufferedReader reader;

    private StringTokenizer tokenizer;

    public FastScanner(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
    }

    public int[] nextIntArray(int length) {
        int[] array = new int[length];
        for (int i = 0; i < length; i++)
            array[i] = nextInt();

        return array;
    }

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

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(nextLine());

        return tokenizer.nextToken();
    }

    public String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

}