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;
const int MAX = 5e5, MAX2 = 1e6;
int n, k, t[MAX + 1], d[MAX2 + 1], cur;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) cin >> t[i];
  int l_ans = 1, r_ans = 1, j = 1, ans = 1;
  for (int i = 1; i < n + 1; i++) {
    int x = t[i];
    if (d[x] < j) ++cur;
    d[x] = i;
    while (cur > k) {
      int y = t[j];
      if (d[y] == j) --cur;
      ++j;
    }
    if (ans < i - j + 1) {
      ans = i - j + 1;
      l_ans = j;
      r_ans = i;
    }
  }
  cout << l_ans << " " << r_ans << '\n';
  return 0;
}
