#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
int n, m;
int a[MAXN], visit[MAXN];
int main() {
  int n, k;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l = 1, r = 1, count_ = 0, max_n = 0, index1 = 0, index2 = 0;
  for (int r = 1; r <= n; r++) {
    visit[a[r]]++;
    if (visit[a[r]] == 1) count_++;
    while (count_ > m) {
      visit[a[l]]--;
      if (visit[a[l]] == 0) count_--;
      l++;
    }
    if (r - l + 1 > max_n) {
      max_n = r - l + 1;
      index1 = l;
      index2 = r;
    }
  }
  printf("%d %d\n", index1, index2);
}

### 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;
const int MAXN = 1e6 + 5;
int n, m;
int a[MAXN], visit[MAXN];
int main() {
  int n, k;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l = 1, r = 1, count_ = 0, max_n = 0, index1 = 0, index2 = 0;
  for (int r = 1; r <= n; r++) {
    visit[a[r]]++;
    if (visit[a[r]] == 1) count_++;
    while (count_ > m) {
      visit[a[l]]--;
      if (visit[a[l]] == 0) count_--;
      l++;
    }
    if (r - l + 1 > max_n) {
      max_n = r - l + 1;
      index1 = l;
      index2 = r;
    }
  }
  printf("%d %d\n", index1, index2);
}

```