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 main() {
  int n, k;
  std::vector<int> v;
  std::cin >> n >> k;
  v.reserve(n);
  for (int i = 0; i < n; i++) {
    int e;
    std::cin >> e;
    v.push_back(e);
  }
  int l = 0, r = 0, max_l = 0, max_r = 0, uniques = 0;
  std::vector<int> freq(1000001, 0);
  while (r < n) {
    if (uniques <= k) {
      if (freq[v[r]] == 0) {
        uniques++;
      }
      freq[v[r]]++;
      if (uniques <= k && r - l > max_r - max_l) {
        max_l = l;
        max_r = r;
      }
      r++;
    } else {
      if (freq[v[l]] == 1) {
        uniques--;
      }
      freq[v[l]]--;
      l++;
    }
  }
  std::cout << max_l + 1 << " " << max_r + 1;
  return 0;
}
