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 N;
  scanf("%d", &N);
  map<int, int> table;
  vector<pair<int, int> > ans;
  for (int i = 0, last = 0; i < N; i++) {
    int p;
    scanf("%d", &p);
    if (++table[p] > 1) {
      ans.emplace_back(last + 1, i + 1);
      last = i + 1;
      table.clear();
    }
  }
  if (!ans.size())
    printf("-1\n");
  else {
    ans.back().second = max(ans.back().second, N);
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++)
      printf("%d %d\n", ans[i].first, ans[i].second);
  }
  return 0;
}
