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 maxA = 1000000;
pair<int, int> getMax(pair<int, int> a, pair<int, int> b) {
  if (a.second - a.first > b.second - b.first)
    return a;
  else
    return b;
}
int main() {
  int n, k, i, a;
  pair<int, int> result(-1, -1);
  cin >> n >> k;
  queue<int> q;
  int fr[maxA + 1] = {}, differentValues = 0;
  for (i = 0; i < n; i++) {
    cin >> a;
    if (!fr[a]) {
      differentValues++;
      q.push(a);
      fr[a]++;
      while (differentValues > k) {
        fr[q.front()]--;
        if (!fr[q.front()]) differentValues--;
        q.pop();
      }
    } else
      q.push(a), fr[a]++;
    result = getMax(result, make_pair(i - ((int)q.size() - 1), i));
  }
  cout << result.first + 1 << ' ' << result.second + 1;
  return 0;
}