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;
using ll = long long;
map<int, int> freq;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int l = 0, r = 0;
  int maxl = 0, maxr = 0, unique = 1;
  freq[a[0]] = 1;
  while (max(l, r) < n) {
    if (unique <= k) {
      if (r - l > maxr - maxl) {
        maxl = l;
        maxr = r;
      }
      r++;
      if (r < n) {
        if (!freq[a[r]]) unique++;
        freq[a[r]]++;
      }
    } else {
      if (freq[a[l]] <= 1) unique--;
      freq[a[l]]--;
      l++;
    }
  }
  printf("%d %d\n", maxl + 1, maxr + 1);
}
