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
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;


public class ProblemB {
    public static void main(String[] args) {
        int testMode = 0;
        String testInput = "5\n" +
                "1 2 2 3 1";
        /*-------------------------------------------------------------------------------------------------------*/
        FastReader in = new FastReader(testMode == 0 ? System.in : new ByteArrayInputStream(testInput.getBytes()));
        PrintWriter out = new PrintWriter(System.out);
        new ProblemSolver().solve(in, out);
        out.close();
    }
}

class ProblemSolver {
    final static int M = 1000001;

    void solve(FastReader in, PrintWriter out) {
        int n = in.nextInt();
        Pair[] a = new Pair[M];
        for (int i = 0; i < M; i++) a[i] = new Pair(0, n, -1);
        for (int i = 0; i < n; i++) {
            Pair p = a[in.nextInt()];
            p.f += 1;
            p.l = Math.min(p.l, i);
            p.r = Math.max(p.r, i);
        }
        Arrays.sort(a);
        out.println((a[0].l + 1) + " " + (a[0].r + 1));
    }
}

class Pair implements Comparable<Pair> {
    public long f;
    public int l;
    public int r;

    public Pair(long f, int l, int r) {
        this.f = f;
        this.l = l;
        this.r = r;
    }

    public int compareTo(Pair o) {
        if (this.f > o.f) return -1;
        if (this.f == o.f) return ((this.r - this.l) - (o.r - o.l));
        return 1;
    }
}

class FastReader extends BufferedReader {
    private StringTokenizer st;

    public FastReader(InputStream is) {
        super(new InputStreamReader(is));
    }

    public String nextToken() {
        try {
            while (st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(readLine());
            }
            return st.nextToken();
        } catch (IOException e) {
            return null;
        }
    }

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

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public int[] nextArrayInt(int size) {
        int[] ret = new int[size];
        for (int i = 0; i < size; i++) {
            ret[i] = nextInt();
        }
        return ret;
    }
}