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
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  scanf("%lli%lli", &n, &k);
  long long arr[n];
  for (long long i = 0; i < n; i++) scanf("%lli", &arr[i]);
  long long start = 0, end = 0;
  long long max_length = 0;
  long long ans_start = -1, ans_end = -1;
  map<long long, long long> hm;
  long long count = 0;
  while (end < n) {
    hm[arr[end]]++;
    if (hm[arr[end]] == 1) count++;
    while (count > k) {
      hm[arr[start]]--;
      if (hm[arr[start]] == 0) count--;
      start++;
    }
    long long length = end - start + 1;
    if (length > max_length) {
      max_length = length;
      ans_start = start;
      ans_end = end;
    }
    end++;
  }
  printf("%lli %lli", ans_start + 1, ans_end + 1);
  return 0;
}
