#include <bits/stdc++.h>
using namespace std;
const int MX = 300005;
long long N, x, y;
int arr[MX];
map<int, int> mp;
pair<int, int> ans[MX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i];
  int j = 0, ok = 0, k = 0;
  for (int i = 0; i < N; i++) {
    mp[arr[i]]++;
    if (mp[arr[i]] > 1) {
      mp.clear();
      ok++;
      ans[k].first = j + 1;
      j = i + 1;
      ans[k].second = i + 1;
      k++;
    }
  }
  if (!ok)
    cout << "-1\n";
  else {
    ans[k - 1].second = N;
    cout << k << "\n";
    for (int i = 0; i < k; i++)
      cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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 MX = 300005;
long long N, x, y;
int arr[MX];
map<int, int> mp;
pair<int, int> ans[MX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i];
  int j = 0, ok = 0, k = 0;
  for (int i = 0; i < N; i++) {
    mp[arr[i]]++;
    if (mp[arr[i]] > 1) {
      mp.clear();
      ok++;
      ans[k].first = j + 1;
      j = i + 1;
      ans[k].second = i + 1;
      k++;
    }
  }
  if (!ok)
    cout << "-1\n";
  else {
    ans[k - 1].second = N;
    cout << k << "\n";
    for (int i = 0; i < k; i++)
      cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}

```