#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int a[n + 1], start[n + 1], end[n + 1], from = 0, cnt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  set<int> used;
  for (int i = 0; i < n; ++i) {
    if (used.find(a[i]) != used.end()) {
      start[cnt] = from;
      used.clear();
      end[cnt] = i;
      cnt++;
      from = i + 1;
    } else {
      used.insert(a[i]);
    }
  }
  end[cnt - 1] = n - 1;
  if (cnt == 0) {
    cout << "-1";
    return 0;
  } else {
    cout << cnt << endl;
    for (int i = 0; i < cnt; ++i) {
      cout << start[i] + 1 << " " << end[i] + 1 << endl;
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 main() {
  int n;
  cin >> n;
  int a[n + 1], start[n + 1], end[n + 1], from = 0, cnt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  set<int> used;
  for (int i = 0; i < n; ++i) {
    if (used.find(a[i]) != used.end()) {
      start[cnt] = from;
      used.clear();
      end[cnt] = i;
      cnt++;
      from = i + 1;
    } else {
      used.insert(a[i]);
    }
  }
  end[cnt - 1] = n - 1;
  if (cnt == 0) {
    cout << "-1";
    return 0;
  } else {
    cout << cnt << endl;
    for (int i = 0; i < cnt; ++i) {
      cout << start[i] + 1 << " " << end[i] + 1 << endl;
    }
  }
}

```