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;
int main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  set<int> st;
  vector<int> vec;
  int start = 1, end = 1;
  for (int i = 0; i < n; i++) {
    if (st.count(arr[i]) == 0) {
      st.insert(arr[i]);
      end++;
    } else {
      vec.push_back(start);
      vec.push_back(end);
      start = end + 1;
      end = start;
      st.clear();
    }
  }
  if (vec.empty()) {
    cout << "-1" << endl;
  } else {
    cout << (vec.size() / 2) << endl;
    vec.pop_back();
    vec.push_back(n);
    auto it = vec.begin();
    while (it != vec.end()) {
      cout << *it << " ";
      it++;
      cout << *it;
      it++;
      cout << endl;
    }
  }
}
