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);
  int n;
  cin >> n;
  vector<pair<int, int> > ans;
  set<int> acc;
  int a;
  int cnt = 0;
  int last = 1;
  bool exist = false;
  bool badlast = false;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    if (acc.find(a) != acc.end()) {
      ++cnt;
      acc.clear();
      ans.push_back(make_pair(last, i + 1));
      last = i + 2;
      exist = true;
      badlast = false;
    } else {
      acc.insert(a);
      badlast = true;
    }
  }
  if (!exist) {
    cout << -1 << "\n";
  } else {
    if (badlast) {
      ans[ans.size() - 1].second = n;
    }
    cout << cnt << "\n";
    for (int i = 0; i < ans.size(); ++i) {
      cout << ans[i].first << " " << ans[i].second << "\n";
    }
  }
}