#include <bits/stdc++.h>
using namespace std;
const int M = 1000 * 1000;
vector<pair<int, int> > ans;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  set<int> s;
  s.clear();
  int st = 0;
  int i = 0;
  while (true) {
    if (i == n) break;
    int tmp;
    cin >> tmp;
    int sz = s.size();
    s.insert(tmp);
    if (sz == s.size()) {
      ans.push_back(make_pair(st + 1, i + 1));
      s.clear();
      st = i + 1;
    }
    i++;
  }
  if (ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  ans[ans.size() - 1].second = n;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  return 0;
}

### 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;
const int M = 1000 * 1000;
vector<pair<int, int> > ans;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  set<int> s;
  s.clear();
  int st = 0;
  int i = 0;
  while (true) {
    if (i == n) break;
    int tmp;
    cin >> tmp;
    int sz = s.size();
    s.insert(tmp);
    if (sz == s.size()) {
      ans.push_back(make_pair(st + 1, i + 1));
      s.clear();
      st = i + 1;
    }
    i++;
  }
  if (ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  ans[ans.size() - 1].second = n;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  return 0;
}

```