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;
map<int, int> mp;
int n, k;
int j = 1;
int arr[500005];
int d[3] = {0};
int main() {
  scanf("%d%d", &n, &k);
  for (int g = 1; g <= n; ++g) scanf("%d", arr + g);
  for (int i = 1; i <= n; ++i) {
    mp[arr[i]]++;
    if (mp.size() <= k) {
      if ((i - j) + 1 > d[0]) {
        d[0] = (i - j) + 1;
        d[1] = j;
        d[2] = i;
      }
    } else if (mp.size() > k) {
      if (mp[arr[j]] > 1)
        mp[arr[j]]--;
      else
        mp.erase(arr[j]);
      ++j;
    }
  }
  printf("%d %d", d[1], d[2]);
  return 0;
}