#include <bits/stdc++.h>
using namespace std;
int cnt[1000100], a[1000100];
int main() {
  int n, k, ans;
  scanf("%d", &n);
  scanf("%d", &k);
  for (long long i = 0; i < n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0, al, ar;
  set<int> x;
  cnt[a[0]]++;
  x.insert(a[0]);
  ans = 1;
  al = 0, ar = 0;
  while (l < n && r < n) {
    r++;
    x.insert(a[r]);
    cnt[a[r]]++;
    while ((long long)x.size() > k && l <= r) {
      cnt[a[l]]--;
      if (cnt[a[l]] == 0) x.erase(a[l]);
      l++;
    }
    r = max(l, r);
    if (r >= n) break;
    if (r - l + 1 > ans) {
      al = l;
      ar = r;
      ans = r - l + 1;
    }
  }
  printf("%d %d\n", al + 1, ar + 1);
  return 0;
}

### 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 cnt[1000100], a[1000100];
int main() {
  int n, k, ans;
  scanf("%d", &n);
  scanf("%d", &k);
  for (long long i = 0; i < n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0, al, ar;
  set<int> x;
  cnt[a[0]]++;
  x.insert(a[0]);
  ans = 1;
  al = 0, ar = 0;
  while (l < n && r < n) {
    r++;
    x.insert(a[r]);
    cnt[a[r]]++;
    while ((long long)x.size() > k && l <= r) {
      cnt[a[l]]--;
      if (cnt[a[l]] == 0) x.erase(a[l]);
      l++;
    }
    r = max(l, r);
    if (r >= n) break;
    if (r - l + 1 > ans) {
      al = l;
      ar = r;
      ans = r - l + 1;
    }
  }
  printf("%d %d\n", al + 1, ar + 1);
  return 0;
}

```