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;
set<int> present;
vector<pair<int, int> > res;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, a;
  cin >> n;
  int p = 1;
  for (int i = 0; i < int(n); i++) {
    cin >> a;
    if (present.find(a) != present.end()) {
      res.push_back(make_pair(p, i + 1));
      p = i + 2;
      present.clear();
      continue;
    }
    present.insert(a);
  }
  if (((int)(res).size()) > 0) {
    res[((int)(res).size()) - 1].second = n;
    cout << ((int)(res).size()) << '\n';
    ;
    for (int i = 0; i < int(((int)(res).size())); i++)
      cout << res[i].first << " " << res[i].second << '\n';
    ;
  } else
    cout << -1 << '\n';
  ;
  return 0;
}