#include <bits/stdc++.h>
using namespace std;
const int MAX = 5 * 1e5;
int a[MAX];
int mp[1000001];
int le, ri, cur = -1, st;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  int cursize = 0;
  for (int i = 0; i < n; i++) {
    if (++mp[a[i]] == 1) cursize++;
    while (cursize > k) {
      if (--mp[a[st]] == 0) {
        cursize--;
      }
      st++;
    }
    if (i - st > cur) {
      cur = i - st;
      le = st + 1;
      ri = i + 1;
    }
  }
  cout << le << " " << ri << "\n";
  return 0;
}

### Prompt

Generate a cpp solution to 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 MAX = 5 * 1e5;
int a[MAX];
int mp[1000001];
int le, ri, cur = -1, st;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  int cursize = 0;
  for (int i = 0; i < n; i++) {
    if (++mp[a[i]] == 1) cursize++;
    while (cursize > k) {
      if (--mp[a[st]] == 0) {
        cursize--;
      }
      st++;
    }
    if (i - st > cur) {
      cur = i - st;
      le = st + 1;
      ri = i + 1;
    }
  }
  cout << le << " " << ri << "\n";
  return 0;
}

```