#include <bits/stdc++.h>
using namespace std;
map<int, int> m;
int main() {
  int n;
  scanf("%d", &n);
  int ans[n / 2 + 1], ans1[n / 2 + 1], p = 0;
  int pr = 0;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (m[x] == 0 || m[x] <= pr)
      m[x] = i + 1;
    else if (m[x] != 0) {
      ans[p] = min(pr + 1, m[x]);
      ans1[p++] = i + 1;
      m[x] = 0;
      pr = i + 1;
    }
  }
  if (p == 0)
    printf("-1\n");
  else {
    ans1[p - 1] = max(n, ans1[p - 1]);
    printf("%d\n", p);
    for (int i = 0; i < p; i++) printf("%d %d\n", ans[i], ans1[i]);
  }
  return 0;
}

### Prompt

Your task is to create 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;
map<int, int> m;
int main() {
  int n;
  scanf("%d", &n);
  int ans[n / 2 + 1], ans1[n / 2 + 1], p = 0;
  int pr = 0;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (m[x] == 0 || m[x] <= pr)
      m[x] = i + 1;
    else if (m[x] != 0) {
      ans[p] = min(pr + 1, m[x]);
      ans1[p++] = i + 1;
      m[x] = 0;
      pr = i + 1;
    }
  }
  if (p == 0)
    printf("-1\n");
  else {
    ans1[p - 1] = max(n, ans1[p - 1]);
    printf("%d\n", p);
    for (int i = 0; i < p; i++) printf("%d %d\n", ans[i], ans1[i]);
  }
  return 0;
}

```