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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, k, c, arr[500005], l, r, diff, ans, freq[1000005];
int diff_num(int l, int r) {
  c = 0;
  for (int i = l; i <= r; i++) {
    if (freq[i] > 0) c++;
  }
  return c;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }
  int l_ans = 0, r_ans = 0;
  r = 0, l = 0, ans = 0, diff = 0;
  while (r != n) {
    freq[arr[r]]++;
    if (freq[arr[r]] == 1) c++;
    if (c <= k) {
      ans = max(ans, r - l + 1);
      if (ans == r - l + 1) {
        r_ans = r + 1;
        l_ans = l + 1;
      }
    }
    while (c > k) {
      freq[arr[l]]--;
      if (freq[arr[l]] == 0) c--;
      l++;
    }
    r++;
  }
  cout << l_ans << " " << r_ans;
}