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;
const long long N = 1e6 + 10;
long long a[N];
set<long long> s;
multiset<long long> ms;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  pair<long long, long long> ans;
  ans = {0, 0};
  long long start = 1;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    s.insert(a[i]);
    ms.insert(a[i]);
    if (s.size() > k) {
      while (s.size() > k) {
        ms.erase(ms.find(a[start]));
        if (ms.find(a[start]) == ms.end()) s.erase(a[start]);
        start++;
      }
    }
    if (s.size() <= k) {
      if (i - start >= ans.second - ans.first) {
        ans = {start, i};
      }
    }
  }
  cout << ans.first << " " << ans.second;
  return 0;
}
