#include <bits/stdc++.h>
using namespace std;
int n, k, a[500005], ans, ansl, ansr, l, g[1000005], c[1000005], s;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  l = 1;
  for (int i = 1; i <= n; i++) {
    if (i - l > ans) {
      ans = i - l;
      ansl = l;
      ansr = i - 1;
    }
    g[a[i]] = i, c[a[i]]++;
    if (c[a[i]] == 1) s++;
    while (s > k) {
      c[a[l]]--;
      if (!c[a[l++]]) s--;
    }
  }
  if (n + 1 - l > ans) {
    ansl = l;
    ansr = n;
  }
  printf("%d %d\n", ansl, ansr);
  return 0;
}

### Prompt

In CPP, your task is to solve 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, a[500005], ans, ansl, ansr, l, g[1000005], c[1000005], s;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  l = 1;
  for (int i = 1; i <= n; i++) {
    if (i - l > ans) {
      ans = i - l;
      ansl = l;
      ansr = i - 1;
    }
    g[a[i]] = i, c[a[i]]++;
    if (c[a[i]] == 1) s++;
    while (s > k) {
      c[a[l]]--;
      if (!c[a[l++]]) s--;
    }
  }
  if (n + 1 - l > ans) {
    ansl = l;
    ansr = n;
  }
  printf("%d %d\n", ansl, ansr);
  return 0;
}

```