Problem: The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Solution {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    static final int Mod = 1000000007;

    static class Pair {
        int x, y;

        Pair(int _x, int _y) {
            x = _x;
            y = _y;
        }


    }

    void solve() throws IOException {
        int n = nextInt();
        int a[] = new int[n];
        Pair p[] = new Pair[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            p[i] = new Pair(a[i], i);
        }
        Arrays.sort(p, new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                if (o1.x == o2.x) return Integer.compare(o1.y, o2.y);
                return Integer.compare(o1.x, o2.x);
            }
        });
        int t = 0;
        for (int i = 0; i < n; i++) {
            if (i == 0 || p[i].x != p[i - 1].x) t++;
            a[p[i].y] = t;
        }
        int cnt[] = new int[350010];
        int i, j;
        Pair last = new Pair(-1, -1);
        List<Pair> ans = new ArrayList<Pair>();
        for (i = 0, j = 0; i < n; i = j + 1) {
            boolean ok = false;
            for (j = i; j < n; j++) {
                if (cnt[a[j]] == 1) {
                    ok = true;
                    break;
                }
                cnt[a[j]] = 1;
            }
            if (ok) ans.add(last = new Pair(i, j));
            for (int k = i; k < n && k <= j; k++) cnt[a[k]] = 0;
        }
        if (ans.size() == 0) {
            out.println(-1);
            return;
        }

        out.println(ans.size());
        if (i != n) {
            ans.remove(last);
            last.y = n - 1;
            ans.add(last);
        }

        for (Pair o : ans) {
            out.println((o.x + 1) + " " + (o.y + 1));
        }
    }

    Solution() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Solution();
    }

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

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

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

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