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 n, k, arr[500009], freq[1000009], l, r, l1, mx;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; ++i) {
    if (freq[arr[i]])
      ++freq[arr[i]];
    else if (k) {
      freq[arr[i]]++;
      --k;
    } else {
      if (i - l1 > mx) {
        mx = i - l1;
        l = l1 + 1;
        r = i;
      }
      while (!k) {
        --freq[arr[l1]];
        if (freq[arr[l1]] == 0) {
          ++k;
          break;
        }
        ++l1;
      }
      ++l1;
      freq[arr[i]]++;
      --k;
    }
  }
  if (n - l1 > mx) {
    l = l1 + 1;
    r = n;
  }
  cout << l << " " << r;
  return 0;
}
