#include <bits/stdc++.h>
using namespace std;
int n;
int a[301309], res;
pair<int, int> luu[301309];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int c = 1;
  for (int i = 1; i <= n; i++) {
    if (m[a[i]] == NULL)
      m[a[i]] = c;
    else {
      luu[++res] = make_pair(c, i);
      c = i + 1;
      m.clear();
    }
    if (i == n) luu[res].second = n;
  }
  if (!res) res--;
  cout << res << endl;
  for (int i = 1; i <= res; i++)
    cout << luu[i].first << " " << luu[i].second << endl;
  return 0;
}

### Prompt

Please formulate 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;
int n;
int a[301309], res;
pair<int, int> luu[301309];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int c = 1;
  for (int i = 1; i <= n; i++) {
    if (m[a[i]] == NULL)
      m[a[i]] = c;
    else {
      luu[++res] = make_pair(c, i);
      c = i + 1;
      m.clear();
    }
    if (i == n) luu[res].second = n;
  }
  if (!res) res--;
  cout << res << endl;
  for (int i = 1; i <= res; i++)
    cout << luu[i].first << " " << luu[i].second << endl;
  return 0;
}

```