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;
void Suhaib_Sawalha() {
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  pair<int, int> ans;
  int longest = 0, l = 0, r = 0;
  map<int, int> m;
  m[a[0]]++;
  while (r < n) {
    if (m.size() <= k) {
      if (r - l + 1 > longest) {
        longest = r - l + 1;
        ans = {l + 1, r + 1};
      }
      r++;
      m[a[r]]++;
    } else {
      m[a[l]]--;
      if (!m[a[l]]) m.erase(a[l]);
      l++;
    }
  }
  cout << ans.first << ' ' << ans.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int _;
  Suhaib_Sawalha();
  return 0;
}