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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  set<long long> s;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  if ((long long)(s.size()) == n) {
    cout << -1 << '\n';
    return;
  }
  map<long long, long long> has;
  vector<pair<long long, long long>> ans;
  long long lasti = 0;
  for (long long i = 0; i < n; i++) {
    if (!has.count(a[i])) {
      has[a[i]] = i;
    } else {
      ans.push_back({min(lasti + 1, has[a[i]] + 1), i + 1});
      lasti = i + 1;
      has.clear();
    }
  }
  long long N = (long long)(ans.size());
  ans[N - 1].second = n;
  cout << (long long)(ans.size()) << '\n';
  for (long long i = 0; i < N; i++) {
    cout << ans[i].first << " " << ans[i].second << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}