#include <bits/stdc++.h>
using namespace std;
set<int> s;
int n;
vector<int> ans;
bool have;
int main() {
  cin >> n;
  for (int a, i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (!s.count(a)) {
      s.insert(a);
    } else {
      s.clear();
      ans.push_back(i + 1);
      have = 1;
    }
  }
  if (!have)
    cout << -1;
  else {
    int p = 0;
    if (!s.empty())
      p = 1;
    else
      p = 1;
    printf("%d\n", ans.size() - p + 1);
    cout << 1 << " ";
    for (int i = 0; i < ans.size() - p; i++) {
      printf("%d\n", ans[i] - 1);
      printf("%d ", ans[i]);
    }
    cout << n << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
set<int> s;
int n;
vector<int> ans;
bool have;
int main() {
  cin >> n;
  for (int a, i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (!s.count(a)) {
      s.insert(a);
    } else {
      s.clear();
      ans.push_back(i + 1);
      have = 1;
    }
  }
  if (!have)
    cout << -1;
  else {
    int p = 0;
    if (!s.empty())
      p = 1;
    else
      p = 1;
    printf("%d\n", ans.size() - p + 1);
    cout << 1 << " ";
    for (int i = 0; i < ans.size() - p; i++) {
      printf("%d\n", ans[i] - 1);
      printf("%d ", ans[i]);
    }
    cout << n << endl;
  }
  return 0;
}

```