#include <bits/stdc++.h>
using namespace std;
int t[300009];
int main() {
  int n;
  int last = 0;
  vector<int> v;
  cin >> n;
  set<int> s;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    t[i] = x;
    if (s.find(x) != s.end()) {
      v.push_back(i);
      s.clear();
    } else {
      s.insert(x);
    }
  }
  if (v.size() == 0) {
    cout << -1;
    return 0;
  }
  v[v.size() - 1] = n - 1;
  cout << v.size() << endl;
  last = 1;
  for (int i = 0; i < v.size(); i++) {
    cout << last << " " << v[i] + 1 << endl;
    last = v[i] + 2;
  }
  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;
int t[300009];
int main() {
  int n;
  int last = 0;
  vector<int> v;
  cin >> n;
  set<int> s;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    t[i] = x;
    if (s.find(x) != s.end()) {
      v.push_back(i);
      s.clear();
    } else {
      s.insert(x);
    }
  }
  if (v.size() == 0) {
    cout << -1;
    return 0;
  }
  v[v.size() - 1] = n - 1;
  cout << v.size() << endl;
  last = 1;
  for (int i = 0; i < v.size(); i++) {
    cout << last << " " << v[i] + 1 << endl;
    last = v[i] + 2;
  }
  return 0;
}

```