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.Closeable;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B558Try implements Closeable {
	BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(System.out);
    StringTokenizer stringTokenizer;
    
    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

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

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

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    public void solve()throws IOException {
    	int n;
    	n=nextInt();
    	//int n=in.nextInt();
		int a[]=new int[n];
		int count[]=new int[1000001];
		int start[]=new int[1000001];
		int end[]=new int[1000001];
		for(int i=0;i<n;i++){
			a[i]=nextInt();
			count[a[i]]++;
		}
		for(int i=0;i<n;i++){
			if(start[a[i]]==0){
				start[a[i]]=i+1;
			}
		}
		for(int i=n-1;i>=0;i--){
			if(end[a[i]]==0){
				end[a[i]]=i+1;
			}
		}
		int ind=-1;
		int max=-1;
		int len=100000000;
		int l=-1;
		int r=-1;
		for(int i=0;i<=1000000;i++){
			max=Math.max(count[i],max);
		}
		for(int i=0;i<=1000000;i++){
			if(count[i]==max){
				max=count[i];
				if((end[i]-start[i])<len && count[i]>0){
					len=end[i]-start[i];
					ind=i;
					l=start[i];
					r=end[i];
				}
			}
		}
		writer.println(l+" "+r);
    }
    public static void main(String args[]) throws IOException{ 
    	try(B558Try c=new B558Try()){
    	c.solve();
    	}
    }
    public void close() throws IOException {
        reader.close();
        writer.close();
    }

}