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;
const int mx = 2e6 + 6;
unordered_map<int, int> m;
map<int, int> msif;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int lans = 1, rans = 1;
  int l = 1;
  long long sif = 0;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    a[i] = x;
    m[x]++;
    if (msif[x] == 1) {
      msif[x] = 0;
      sif--;
    }
    while (m.size() - sif > k) {
      m[a[l]]--;
      if (m[a[l]] == 0) {
        msif[a[l]] = 1;
        sif++;
      }
      ++l;
    }
    if (i - l + 1 > rans - lans + 1) {
      lans = l;
      rans = i;
    }
  }
  cout << lans << " " << rans << '\n';
}