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;
map<int, int> o;
int main() {
  vector<pair<int, int> > res;
  int n;
  cin >> n;
  int a[n + 2];
  int last = 1;
  int endd = 1;
  int i;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (o[a[i]] == 0) {
      o[a[i]] = i;
    } else {
      if (o[a[i]] < last) {
        o[a[i]] = i;
      } else {
        res.push_back(make_pair(last, i));
        o[a[i]] = 0;
        last = i + 1;
      }
    }
  }
  if (res.size() == 0) {
    cout << "-1";
    return 0;
  }
  if (res[res.size() - 1].second != n) {
    res[res.size() - 1].second = n;
  }
  cout << res.size() << endl;
  for (i = 0; i < res.size(); i++) {
    cout << res[i].first << " " << res[i].second << endl;
  }
  return 0;
}
