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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int vis[1000010], ar[1000010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  int limit;
  scanf("%d%d", &n, &limit);
  ;
  int counter = 0, index = 1, mx = 0, f, s;
  for (int i = 1; i < n + 1; i++) {
    ;
    int p;
    scanf("%d", &p);
    ;
    ar[i] = p;
  }
  for (int i = 1; i < n + 1; i++) {
    vis[ar[i]]++;
    if (vis[ar[i]] == 1) counter++;
    if (counter > limit) {
      while (counter > limit) {
        if (vis[ar[index]] == 1) counter--;
        vis[ar[index]]--;
        index++;
      }
    }
    if (i - index + 1 > mx) {
      mx = i - index + 1;
      f = index;
      s = i;
    }
  }
  printf("%d %d\n", f, s);
  return 0;
}