#include <bits/stdc++.h>
int a[1000010];
using namespace std;
set<int> s;
int main() {
  a[1] = 1;
  int n, count = 1, flag = 1;
  scanf("%d", &n);
  int i, m;
  for (i = 1; i <= n; i++) {
    scanf("%d", &m);
    if (s.count(m)) {
      count += 2;
      a[count - 1] = i;
      a[count] = i + 1;
      flag = 0;
      s.clear();
    } else
      s.insert(m);
  }
  if (flag) {
    printf("-1\n");
    return 0;
  }
  count -= 1;
  a[count] = n;
  printf("%d\n", count / 2);
  for (i = 1; i <= count / 2; i++) {
    printf("%d %d\n", a[2 * i - 1], a[2 * i]);
  }
  return 0;
}

### Prompt

Generate 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>
int a[1000010];
using namespace std;
set<int> s;
int main() {
  a[1] = 1;
  int n, count = 1, flag = 1;
  scanf("%d", &n);
  int i, m;
  for (i = 1; i <= n; i++) {
    scanf("%d", &m);
    if (s.count(m)) {
      count += 2;
      a[count - 1] = i;
      a[count] = i + 1;
      flag = 0;
      s.clear();
    } else
      s.insert(m);
  }
  if (flag) {
    printf("-1\n");
    return 0;
  }
  count -= 1;
  a[count] = n;
  printf("%d\n", count / 2);
  for (i = 1; i <= count / 2; i++) {
    printf("%d %d\n", a[2 * i - 1], a[2 * i]);
  }
  return 0;
}

```