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;
int main() {
  int freq[1000100] = {};
  int n, k, num[600100];
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> num[i];
  freq[num[1]]++;
  int ndif = 1, maxlen = 1;
  int l = 1, r = 1;
  int ansl = 1, ansr = 1;
  while (r < n) {
    if (ndif <= k) {
      r++;
      if (freq[num[r]] == 0) ndif++;
      freq[num[r]]++;
    } else {
      l++;
      if (freq[num[l - 1]] == 1) ndif--;
      freq[num[l - 1]]--;
    }
    if (ndif <= k && r - l + 1 > maxlen) {
      maxlen = r - l + 1;
      ansl = l;
      ansr = r;
    }
  }
  cout << ansl << " " << ansr << endl;
  return 0;
}
