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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, i, j, begin, count, l = 0, r = 0;
  cin >> n >> k;
  map<long long int, long long int> m;
  long long int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  begin = 0;
  count = 1;
  m[a[0]] = 1;
  j = 1;
  while (j < n) {
    m[a[j]]++;
    if (m[a[j]] == 1) count++;
    if (count > k) {
      i = begin;
      while (i <= j && m[a[i]] > 1) {
        m[a[i]]--;
        i++;
      }
      m[a[i]]--;
      begin = i + 1;
      count -= 1;
      j++;
      continue;
    }
    if (j - begin > r - l) {
      r = j;
      l = begin;
    }
    j++;
  }
  cout << l + 1 << " " << r + 1;
  return 0;
}