#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 3;
int fr[N];
vector<int> v(N);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  int l = 1;
  int r = 1;
  int mx = -1;
  int ans1 = 1;
  int ans2 = 1;
  int cnt = 0;
  int t = k;
  while (r <= n) {
    if (!fr[v[r]]) {
      cnt++;
    }
    fr[v[r]]++;
    if (cnt <= k) {
      if (r - l + 1 > mx) {
        mx = r - l + 1;
        ans1 = l;
        ans2 = r;
      }
    }
    while (cnt > k) {
      fr[v[l]]--;
      if (fr[v[l]] == 0) {
        cnt--;
      }
      l++;
    }
    r++;
  }
  cout << ans1 << " " << ans2 << "\n";
  return 0;
}

### 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;
const int N = 1e6 + 3;
int fr[N];
vector<int> v(N);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
  }
  int l = 1;
  int r = 1;
  int mx = -1;
  int ans1 = 1;
  int ans2 = 1;
  int cnt = 0;
  int t = k;
  while (r <= n) {
    if (!fr[v[r]]) {
      cnt++;
    }
    fr[v[r]]++;
    if (cnt <= k) {
      if (r - l + 1 > mx) {
        mx = r - l + 1;
        ans1 = l;
        ans2 = r;
      }
    }
    while (cnt > k) {
      fr[v[l]]--;
      if (fr[v[l]] == 0) {
        cnt--;
      }
      l++;
    }
    r++;
  }
  cout << ans1 << " " << ans2 << "\n";
  return 0;
}

```