#include <bits/stdc++.h>
using namespace std;
int a[1000010];
int vis[1000010];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int p1 = 0, p2 = 0, l = 0, r = 0, sum = 0;
  while (p1 < n && p2 < n) {
    if (sum < k || (sum == k && vis[a[p2]] != 0)) {
      if (p2 - p1 >= r - l) l = p1, r = p2;
      if (++vis[a[p2]] == 1) sum++;
      p2++;
    } else {
      if (--vis[a[p1]] == 0) sum--;
      p1++;
    }
  }
  cout << l + 1 << " " << r + 1 << endl;
  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;
int a[1000010];
int vis[1000010];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int p1 = 0, p2 = 0, l = 0, r = 0, sum = 0;
  while (p1 < n && p2 < n) {
    if (sum < k || (sum == k && vis[a[p2]] != 0)) {
      if (p2 - p1 >= r - l) l = p1, r = p2;
      if (++vis[a[p2]] == 1) sum++;
      p2++;
    } else {
      if (--vis[a[p1]] == 0) sum--;
      p1++;
    }
  }
  cout << l + 1 << " " << r + 1 << endl;
  return 0;
}

```