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 int N = 1e6 + 5;
long long int n, k, la, ra, sa;
vector<long long int> v(N);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> k;
  for (long long int i = 0; i <= n - 1; i++) cin >> v[i];
  long long int l = 0, r = 0;
  map<long long int, long long int> m;
  m[v[0]]++;
  while (l < n && r < n) {
    if (m.size() <= k) {
      if (r - l > sa) la = l, ra = r, sa = ra - la;
      r++;
      m[v[r]]++;
    } else {
      m[v[l]]--;
      if (m[v[l]] == 0) {
        m.erase(m.find(v[l]));
      }
      l++;
    }
  }
  cout << la + 1 << " " << ra + 1 << "\n";
}
