#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long a[n], i, m = 0, s = 0, e, len = 0, start, end;
  for (i = 0; i < n; i++) cin >> a[i];
  map<long long, long long> mp;
  for (e = 0; e < n; e++) {
    mp[a[e]]++;
    while (mp.size() > k && s < n) {
      mp[a[s]]--;
      if (mp[a[s]] <= 0) {
        mp.erase(a[s]);
      }
      s++;
    }
    len = e - s + 1;
    if (len > m) {
      m = len;
      start = s;
      end = e;
    }
  }
  cout << start + 1 << " " << end + 1 << "\n";
}

### Prompt

Create a solution in cpp for 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);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long a[n], i, m = 0, s = 0, e, len = 0, start, end;
  for (i = 0; i < n; i++) cin >> a[i];
  map<long long, long long> mp;
  for (e = 0; e < n; e++) {
    mp[a[e]]++;
    while (mp.size() > k && s < n) {
      mp[a[s]]--;
      if (mp[a[s]] <= 0) {
        mp.erase(a[s]);
      }
      s++;
    }
    len = e - s + 1;
    if (len > m) {
      m = len;
      start = s;
      end = e;
    }
  }
  cout << start + 1 << " " << end + 1 << "\n";
}

```