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>
using namespace std;
long long a[1000000];
int temp[1000009] = {0};
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int st, en, c = 0, max_len = 0, ans_st, ans_en;
  st = 1;
  en = 1;
  int i = 1;
  int p = 1;
  while (i <= n) {
    st = i;
    while (p <= n) {
      if (temp[a[p]] == 0) {
        temp[a[p]] = 1;
        c++;
        if (c > k) {
          temp[a[p]]--;
          c--;
          break;
        }
      } else
        temp[a[p]]++;
      p++;
    }
    en = p - 1;
    if (en - st + 1 > max_len) {
      max_len = en - st + 1;
      ans_en = en;
      ans_st = st;
    }
    temp[a[i]]--;
    if (temp[a[i]] == 0) c--;
    i++;
  }
  printf("%d %d", ans_st, ans_en);
  return 0;
}
