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 INF = 1e18 + 9;
const int N = 2e5 + 5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  int i = 0, j = 0, ans = 0;
  set<int> st;
  multiset<int> ms;
  int a, b;
  while (j < n) {
    if (int(st.size()) < k || (int(st.size()) == k && st.count(v[j]))) {
      st.insert(v[j]);
      ms.insert(v[j]);
      j++;
      if (j - i > ans) {
        ans = j - i;
        a = i + 1;
        b = j;
      }
    } else {
      while (i < j) {
        ms.erase(ms.find(v[i]));
        if (ms.find(v[i]) == ms.end()) {
          st.erase(v[i]);
          break;
        }
        i++;
      }
      i++;
    }
  }
  cout << a << " " << b << endl;
  return 0;
}
