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;
int v[5000010], f[1000010], n, k, nr, maxi, start, stop, lung;
deque<int> d;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    d.push_back(i);
    if (f[v[i]] == 0) {
      f[v[i]] = 1;
      nr++;
    } else {
      f[v[i]]++;
    }
    lung++;
    while (nr > k) {
      int a = d.front();
      d.pop_front();
      f[v[a]]--;
      lung--;
      if (f[v[a]] == 0) --nr;
    }
    if (lung > maxi) {
      maxi = lung;
      start = d.front();
      stop = d.back();
    }
  }
  cout << start << " " << stop;
  return 0;
}
