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.*;
import java.util.*;

public class Test {
    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        // InputStream inputStream = new FileInputStream(new File("stars.in"));
        // OutputStream outputStream = new FileOutputStream(new
        // File("stars.out"));

        Input in = new Input(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task task = new Task();
        task.solve(in, out);
        out.close();
    }
}

class Input {
    public BufferedReader bufferedReader;
    private StringTokenizer stringTokenizer;

    public Input(InputStream inputStream) {
        bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        stringTokenizer = null;
    }

    public String nextString() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        }
        return stringTokenizer.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextString());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }

    public int[] nextArrayInt(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }
}

class Task {
    public void solve(Input in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int[] b = in.nextArrayInt(n);

        int[] a = new int[1000001];
        int[] l = new int[1000001];
        int[] r = new int[1000001];
        boolean[] flag = new boolean[1000001];
        
        for (int i = 0; i < n; i++) {
            if (!flag[b[i]]) {
                l[b[i]] = i + 1;
                flag[b[i]] = true;
            } else {
                r[b[i]] = i + 1;
            }
            a[b[i]]++;
        }

        int x = 0;
        for (int i = 1; i < a.length; i++) {
            if (x < a[i]) {
                x = a[i];
            }
        }
        
        if(x == 1){
            out.println("1 1");
            return;
        }
        
        int ansl = 0;
        int ansr = n;
        
        for (int i = 1; i < a.length; i++){
            if(a[i] == x && ansr-ansl > r[i]-l[i]){
                ansl = l[i];
                ansr = r[i];
            }
        }
        
        out.println(ansl + " " + ansr);
    }
}