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 main() {
  int n, k;
  cin >> n >> k;
  int maxx = 2000000;
  vector<int> frequency(maxx, 0);
  vector<int> val(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> val[i];
  }
  int left = 1;
  int right = 1;
  int distinct = 0;
  int best_left = 1;
  int best_right = 1;
  int best_distance = 0;
  while (right <= n) {
    frequency[val[right]]++;
    if (frequency[val[right]] == 1) distinct++;
    while (distinct > k) {
      frequency[val[left]]--;
      if (frequency[val[left]] == 0) distinct--;
      left++;
    }
    if ((right - left) + 1 > best_distance) {
      best_left = left;
      best_right = right;
      best_distance = (right - left) + 1;
    }
    right++;
  }
  cout << best_left << " " << best_right;
  return 0;
}
