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;
const int N = 2e6 + 1;
int n;
set<int> st;
vector<pair<int, int> > ans;
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> n;
  int l = 1;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if ((int)st.size() && st.find(x) != st.end()) {
      st.clear();
      ans.push_back({l, i});
      l = i + 1;
      continue;
    }
    st.insert(x);
  }
  if (!(int)ans.size()) {
    cout << -1;
    return 0;
  }
  if ((int)st.size()) {
    ans.back().second = n;
  }
  cout << (int)ans.size() << '\n';
  for (auto it : ans) {
    cout << it.first << ' ' << it.second << '\n';
  }
  return 0;
}