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;
long long int power(int b, int pw) {
  long long int res = 1;
  for (int i = 1; i <= pw; i++) {
    res = res * b;
  }
  return res;
}
long long int typ[1000000];
int main() {
  ios_base::sync_with_stdio(false);
  long long int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> typ[i];
  }
  set<int> prev;
  vector<pair<int, int> > ns;
  long long int cnt = 0;
  long long int from = 1;
  for (int i = 1; i <= n; i++) {
    if (prev.find(typ[i]) != prev.end()) {
      ns.push_back(make_pair(from, i));
      prev.clear();
      from = i + 1;
    } else {
      prev.insert(typ[i]);
    }
  }
  if (ns.size() == 0) {
    printf("-1\n");
    return 0;
  }
  cout << ns.size() << endl;
  int temp = ns.size();
  for (int i = 0; i < ns.size(); i++) {
    if (i == temp - 1) {
      cout << ns[i].first << " " << n << endl;
      return 0;
    }
    cout << ns[i].first << " " << ns[i].second << endl;
  }
  return 0;
}
