#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
const int Q = 1e9 + 7;
int n, K, a[N], num[N], cnt;
void work() {
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  int res = -1 << 30, E;
  for (int i = 0, j = -1; i < n; ++i) {
    if (++num[a[i]] == 1) ++cnt;
    while (cnt > K) {
      ++j;
      if (--num[a[j]] == 0) --cnt;
    }
    if (i - j > res) {
      res = i - j;
      E = i;
    }
  }
  printf("%d %d\n", E - res + 2, E + 1);
}
int main() {
  work();
  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;
const int N = 1000005;
const int Q = 1e9 + 7;
int n, K, a[N], num[N], cnt;
void work() {
  scanf("%d%d", &n, &K);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  int res = -1 << 30, E;
  for (int i = 0, j = -1; i < n; ++i) {
    if (++num[a[i]] == 1) ++cnt;
    while (cnt > K) {
      ++j;
      if (--num[a[j]] == 0) --cnt;
    }
    if (i - j > res) {
      res = i - j;
      E = i;
    }
  }
  printf("%d %d\n", E - res + 2, E + 1);
}
int main() {
  work();
  return 0;
}

```