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
#include <bits/stdc++.h>
int brojevi[1000000] = {0};
int main() {
  int a[500002];
  int tmp, n, k, i, j, razliciti = 0;
  int max_i = 0, max_j = 0, max_diff = 0;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  i = 0;
  j = 0;
  while (j < n) {
    if (brojevi[a[j]]) {
      brojevi[a[j]]++;
    } else {
      brojevi[a[j]] = 1;
      razliciti++;
      while (razliciti > k) {
        if (brojevi[a[i]] == 1) {
          brojevi[a[i]] = 0;
          razliciti--;
          i++;
        } else {
          brojevi[a[i]]--;
          i++;
        }
      }
    }
    if (max_diff < j - i + 1) {
      max_diff = j - i + 1;
      max_i = i;
      max_j = j;
    }
    j++;
  }
  printf("%d %d\n", max_i + 1, max_j + 1);
  return 0;
}
