#include <bits/stdc++.h>
using namespace std;
int num[300005];
map<int, int> p;
int main() {
  int n;
  cin >> n;
  int k;
  int ans = 0, j = 0, l = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &k);
    p[k] += 1;
    if (p[k] == 2) {
      ans++;
      j += 2;
      num[j - 2] = l;
      num[j - 1] = i;
      l = i + 1;
      p.clear();
    }
  }
  if (ans == 0) {
    cout << "-1" << endl;
    return 0;
  }
  num[j - 1] = n;
  cout << ans << endl;
  for (int i = 0; i < j; i += 2) printf("%d %d\n", num[i], num[i + 1]);
  return 0;
}

### 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;
int num[300005];
map<int, int> p;
int main() {
  int n;
  cin >> n;
  int k;
  int ans = 0, j = 0, l = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &k);
    p[k] += 1;
    if (p[k] == 2) {
      ans++;
      j += 2;
      num[j - 2] = l;
      num[j - 1] = i;
      l = i + 1;
      p.clear();
    }
  }
  if (ans == 0) {
    cout << "-1" << endl;
    return 0;
  }
  num[j - 1] = n;
  cout << ans << endl;
  for (int i = 0; i < j; i += 2) printf("%d %d\n", num[i], num[i + 1]);
  return 0;
}

```