#include <bits/stdc++.h>
using namespace std;
void init() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
int fr[int(1e6) + 1];
int main() {
  init();
  int n, k;
  cin >> n >> k;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int l = 0, r = 0, distinct = 0, ans = 0, ansl = 0, ansr = 0;
  while (r < n) {
    if (!fr[arr[r]]) distinct++;
    fr[arr[r]]++;
    while (distinct > k) {
      fr[arr[l]]--;
      if (!fr[arr[l]]) distinct--;
      l++;
    }
    if (r - l + 1 > ans) {
      ans = r - l + 1;
      ansl = l + 1;
      ansr = r + 1;
    }
    r++;
  }
  cout << ansl << " " << ansr;
}

### 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;
void init() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
int fr[int(1e6) + 1];
int main() {
  init();
  int n, k;
  cin >> n >> k;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  int l = 0, r = 0, distinct = 0, ans = 0, ansl = 0, ansr = 0;
  while (r < n) {
    if (!fr[arr[r]]) distinct++;
    fr[arr[r]]++;
    while (distinct > k) {
      fr[arr[l]]--;
      if (!fr[arr[l]]) distinct--;
      l++;
    }
    if (r - l + 1 > ans) {
      ans = r - l + 1;
      ansl = l + 1;
      ansr = r + 1;
    }
    r++;
  }
  cout << ansl << " " << ansr;
}

```