#include <bits/stdc++.h>
using namespace std;
const int siz = 10 + 1e6;
const int siz1 = 5e5 + 10;
int f[siz], a[siz1], n, k, len, L, R;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0, cnt = 0;
  while (r < n) {
    if (f[a[r]] == 0) cnt++;
    f[a[r]]++;
    if (cnt <= k)
      if (len < r - l + 1) {
        L = l;
        R = r;
        len = r - l + 1;
      }
    while (cnt > k) {
      f[a[l]]--;
      if (f[a[l]] == 0) cnt--;
      l++;
    }
    r++;
  }
  printf("%d %d", ++L, ++R);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 siz = 10 + 1e6;
const int siz1 = 5e5 + 10;
int f[siz], a[siz1], n, k, len, L, R;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0, cnt = 0;
  while (r < n) {
    if (f[a[r]] == 0) cnt++;
    f[a[r]]++;
    if (cnt <= k)
      if (len < r - l + 1) {
        L = l;
        R = r;
        len = r - l + 1;
      }
    while (cnt > k) {
      f[a[l]]--;
      if (f[a[l]] == 0) cnt--;
      l++;
    }
    r++;
  }
  printf("%d %d", ++L, ++R);
  return 0;
}

```