#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  set<int> S;
  cin >> n;
  vector<pair<int, int> > bra;
  int fans = 0;
  int pend = -1;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (S.find(x) != S.end()) {
      bra.push_back(make_pair(pend + 1, i));
      pend = i;
      S.clear();
      continue;
    }
    S.insert(x);
  }
  int sz = (int)bra.size();
  if (!sz) {
    cout << -1 << "\n";
    return 0;
  }
  bra[sz - 1].second = n - 1;
  cout << (int)bra.size() << "\n";
  for (int i = 0; i < bra.size(); ++i) {
    cout << bra[i].first + 1 << " " << bra[i].second + 1 << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  set<int> S;
  cin >> n;
  vector<pair<int, int> > bra;
  int fans = 0;
  int pend = -1;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (S.find(x) != S.end()) {
      bra.push_back(make_pair(pend + 1, i));
      pend = i;
      S.clear();
      continue;
    }
    S.insert(x);
  }
  int sz = (int)bra.size();
  if (!sz) {
    cout << -1 << "\n";
    return 0;
  }
  bra[sz - 1].second = n - 1;
  cout << (int)bra.size() << "\n";
  for (int i = 0; i < bra.size(); ++i) {
    cout << bra[i].first + 1 << " " << bra[i].second + 1 << "\n";
  }
  return 0;
}

```