#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> v;
int freq[1000000] = {};
int main() {
  scanf("%d%d", &n, &k);
  v.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  int l = 0, r = 0, sum = 0;
  int nl = 0, nr = -1;
  while (nr < n) {
    nr++;
    if (nr < n) {
      freq[v[nr]]++;
      if (freq[v[nr]] == 1) sum++;
      while (sum > k) {
        freq[v[nl]]--;
        if (freq[v[nl]] == 0) sum--;
        nl++;
      }
      if (sum <= k && (nr - nl) > (r - l)) {
        r = nr, l = nl;
      }
    }
  }
  printf("%d %d", l + 1, r + 1);
  return 0;
}

### Prompt

Please formulate 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;
int n, k;
vector<int> v;
int freq[1000000] = {};
int main() {
  scanf("%d%d", &n, &k);
  v.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  int l = 0, r = 0, sum = 0;
  int nl = 0, nr = -1;
  while (nr < n) {
    nr++;
    if (nr < n) {
      freq[v[nr]]++;
      if (freq[v[nr]] == 1) sum++;
      while (sum > k) {
        freq[v[nl]]--;
        if (freq[v[nl]] == 0) sum--;
        nl++;
      }
      if (sum <= k && (nr - nl) > (r - l)) {
        r = nr, l = nl;
      }
    }
  }
  printf("%d %d", l + 1, r + 1);
  return 0;
}

```