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;
long long a, x = 0, l = 1, r = 0, n;
vector<pair<int, int> > v;
set<int> s;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (s.find(a) != s.end()) {
      x++;
      r = i;
      v.push_back({l, r});
      l = r + 1;
      while (!s.empty()) s.erase(s.begin());
    } else
      s.insert(a);
  }
  if (x == 0) {
    cout << -1;
    return 0;
  }
  cout << x << '\n';
  for (int i = 0; i < v.size(); i++) {
    if (i == v.size() - 1 && v[i].second != n) {
      v[i].second = n;
    }
    cout << v[i].first << ' ' << v[i].second << '\n';
  }
  return 0;
}