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;
const int MX = 1000000;
set<int> tp;
int n, arr[MX];
vector<pair<int, int> > ans;
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  int ind = 0;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    if (tp.count(arr[i]) > 0) {
      ans.push_back(make_pair(ind, i));
      ind = i + 1;
      tp.clear();
    } else
      tp.insert(arr[i]);
  }
  if ((int)ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  ans[(int)ans.size() - 1].second = n - 1;
  cout << (int)ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); i++) {
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
  }
  return 0;
}
