#include <bits/stdc++.h>
using namespace std;
const int nsz = 3e5;
int n, a[nsz + 5];
vector<pair<int, int> > ans;
map<int, int> cnt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int fm = 1;
  for (int i = 1; i <= n; ++i) {
    ++cnt[a[i]];
    if (cnt[a[i]] == 2) {
      cnt.clear();
      ans.push_back(make_pair(fm, i));
      fm = i + 1;
    }
  }
  if (!ans.size()) {
    cout << -1 << '\n';
    return 0;
  }
  ans.back().second = n;
  cout << ans.size() << '\n';
  for (__typeof(ans.begin()) it = ans.begin(); it != ans.end(); ++it) {
    cout << it->first << ' ' << it->second << '\n';
  }
}

### Prompt

Please provide a Cpp coded solution 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;
const int nsz = 3e5;
int n, a[nsz + 5];
vector<pair<int, int> > ans;
map<int, int> cnt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  int fm = 1;
  for (int i = 1; i <= n; ++i) {
    ++cnt[a[i]];
    if (cnt[a[i]] == 2) {
      cnt.clear();
      ans.push_back(make_pair(fm, i));
      fm = i + 1;
    }
  }
  if (!ans.size()) {
    cout << -1 << '\n';
    return 0;
  }
  ans.back().second = n;
  cout << ans.size() << '\n';
  for (__typeof(ans.begin()) it = ans.begin(); it != ans.end(); ++it) {
    cout << it->first << ' ' << it->second << '\n';
  }
}

```