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 arr[300005];
map<int, int> mp;
vector<int> opt;
int main() {
  int n, lst = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    if (mp[arr[i]] && mp[arr[i]] > lst) {
      mp[arr[i]] = 0;
      opt.push_back(++lst);
      opt.push_back(i);
      lst = i;
    } else {
      mp[arr[i]] = i;
    }
  }
  int l = opt.size();
  if (l / 2 == 0) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", l / 2);
  opt[l - 1] = n;
  for (int i = 0; i < l; i++) {
    printf("%d ", opt[i++]);
    printf("%d\n", opt[i]);
  }
  return 0;
}
