#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 13;
int n, k, a[maxN], cur, st, en, ans, cnt[maxN];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int r = -1;
  for (int i = 0; i < n; i++) {
    if (i) {
      cnt[a[i - 1]]--;
      if (cnt[a[i - 1]] == 0) cur--;
    }
    while (cur <= k && r < n) {
      r++;
      cnt[a[r]]++;
      if (cnt[a[r]] == 1) cur++;
    }
    if (r - i > ans) {
      ans = r - i;
      st = i + 1, en = r;
    }
  }
  cout << st << " " << en << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxN = 1e6 + 13;
int n, k, a[maxN], cur, st, en, ans, cnt[maxN];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int r = -1;
  for (int i = 0; i < n; i++) {
    if (i) {
      cnt[a[i - 1]]--;
      if (cnt[a[i - 1]] == 0) cur--;
    }
    while (cur <= k && r < n) {
      r++;
      cnt[a[r]]++;
      if (cnt[a[r]] == 1) cur++;
    }
    if (r - i > ans) {
      ans = r - i;
      st = i + 1, en = r;
    }
  }
  cout << st << " " << en << endl;
}

```