#include <bits/stdc++.h>
using namespace std;
int a[500005], ck[1000006], ans[500005], yo, cnt, k, n;
int x, y;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (!ck[a[i]]) {
      cnt++;
      if (cnt > k) {
        cnt--;
        ck[a[yo]]--;
        if (!ck[a[yo]]) cnt--;
        yo++;
        i--;
        continue;
      }
    }
    ck[a[i]]++;
    ans[yo] = i;
  }
  int mx = -1;
  for (int i = 0; i < n; i++) {
    int d = ans[i] - i;
    if (d > mx) x = i + 1, y = ans[i] + 1, mx = d;
  }
  printf("%d %d", x, y);
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
int a[500005], ck[1000006], ans[500005], yo, cnt, k, n;
int x, y;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (!ck[a[i]]) {
      cnt++;
      if (cnt > k) {
        cnt--;
        ck[a[yo]]--;
        if (!ck[a[yo]]) cnt--;
        yo++;
        i--;
        continue;
      }
    }
    ck[a[i]]++;
    ans[yo] = i;
  }
  int mx = -1;
  for (int i = 0; i < n; i++) {
    int d = ans[i] - i;
    if (d > mx) x = i + 1, y = ans[i] + 1, mx = d;
  }
  printf("%d %d", x, y);
  return 0;
}

```