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>
long long int mod = 1000000007;
using namespace std;
int main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int ans = 0, index = 0;
  vector<int> answer;
  while (index < n) {
    int flag = 0;
    map<int, int> present;
    int temp = index;
    while (flag == 0 && temp < n) {
      if (present.count(arr[temp]) == 0) {
        present[arr[temp]]++;
        temp++;
      } else {
        flag = 1;
        answer.push_back(index);
        answer.push_back(temp);
      }
    }
    index = temp + 1;
  }
  if (answer.size() == 0) {
    cout << "-1" << endl;
    return 0;
  }
  answer[answer.size() - 1] = n - 1;
  cout << answer.size() / 2 << endl;
  for (int i = 0; i < answer.size(); i += 2) {
    cout << answer[i] + 1 << " " << answer[i + 1] + 1 << endl;
  }
}
