#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k;
  cin >> n >> k;
  int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (k == 1 && a[0] != a[1]) {
    cout << "1 1";
    return 0;
  }
  map<int, int> mp;
  int st = 1, en = 1;
  i = 0;
  j = 1;
  mp[a[i]]++;
  while (j < n) {
    mp[a[j]]++;
    if (mp.size() <= k) {
      if (j - i > en - st) {
        st = i;
        en = j;
      }
    } else if (mp.size() > k) {
      if (mp[a[i]] == 1)
        mp.erase(a[i]);
      else
        mp[a[i]]--;
      i++;
    }
    j++;
  }
  cout << st + 1 << " " << en + 1 << "\n";
}

### Prompt

Generate 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(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, j, k;
  cin >> n >> k;
  int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (k == 1 && a[0] != a[1]) {
    cout << "1 1";
    return 0;
  }
  map<int, int> mp;
  int st = 1, en = 1;
  i = 0;
  j = 1;
  mp[a[i]]++;
  while (j < n) {
    mp[a[j]]++;
    if (mp.size() <= k) {
      if (j - i > en - st) {
        st = i;
        en = j;
      }
    } else if (mp.size() > k) {
      if (mp[a[i]] == 1)
        mp.erase(a[i]);
      else
        mp[a[i]]--;
      i++;
    }
    j++;
  }
  cout << st + 1 << " " << en + 1 << "\n";
}

```