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;
using ll = long long int;
using pll = pair<ll, ll>;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ll n;
  cin >> n;
  ll arr[n];
  for (ll i = 0; i < n; i++) cin >> arr[i];
  set<ll> dup;
  ll l = 0;
  vector<pll> ans;
  for (ll i = 0; i < n; i++) {
    if (dup.find(arr[i]) == dup.end())
      dup.insert(arr[i]);
    else {
      dup.clear();
      ans.push_back({l + 1, i + 1});
      l = i + 1;
    }
  }
  if (ans.size()) {
    ans[((ll)ans.size()) - 1].second = n;
    cout << ans.size() << "\n";
    for (auto it : ans) cout << it.first << " " << it.second << "\n";
  } else
    cout << "-1\n";
  return 0;
}
