#include <bits/stdc++.h>
using namespace std;
int a[1000005];
int vis[1000005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int l = 1;
  int r = 1;
  int cnt = 1;
  int pl = 2;
  vis[a[1]]++;
  for (int i = 1; i <= n; i++) {
    while (pl <= n && cnt <= k) {
      vis[a[pl]]++;
      if (vis[a[pl]] == 1) {
        cnt++;
      }
      if (cnt <= k) {
        pl++;
      }
    }
    if (r - l < pl - 1 - i) {
      l = i;
      r = pl - 1;
    }
    vis[a[i]]--;
    if (vis[a[i]] == 0) {
      cnt--;
      pl++;
    }
  }
  cout << l << " " << r << 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[1000005];
int vis[1000005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int l = 1;
  int r = 1;
  int cnt = 1;
  int pl = 2;
  vis[a[1]]++;
  for (int i = 1; i <= n; i++) {
    while (pl <= n && cnt <= k) {
      vis[a[pl]]++;
      if (vis[a[pl]] == 1) {
        cnt++;
      }
      if (cnt <= k) {
        pl++;
      }
    }
    if (r - l < pl - 1 - i) {
      l = i;
      r = pl - 1;
    }
    vis[a[i]]--;
    if (vis[a[i]] == 0) {
      cnt--;
      pl++;
    }
  }
  cout << l << " " << r << endl;
  return 0;
}

```