#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> ans;
  map<int, bool> mp;
  int last = 1;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    if (mp[t]) {
      ans.push_back({last, i});
      last = i + 1;
      mp.clear();
    } else {
      mp[t] = true;
    }
  }
  if (ans.empty()) {
    cout << "-1\n";
  } else {
    ans.back().second = n;
    cout << ans.size() << '\n';
    for (auto i : ans) {
      cout << i.first << " " << i.second << '\n';
    }
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> ans;
  map<int, bool> mp;
  int last = 1;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    if (mp[t]) {
      ans.push_back({last, i});
      last = i + 1;
      mp.clear();
    } else {
      mp[t] = true;
    }
  }
  if (ans.empty()) {
    cout << "-1\n";
  } else {
    ans.back().second = n;
    cout << ans.size() << '\n';
    for (auto i : ans) {
      cout << i.first << " " << i.second << '\n';
    }
  }
}

```