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;
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int a[500001];
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  map<int, int> already;
  map<int, int> last_seen;
  int left = 1;
  int right = 1;
  int counter = 1;
  already[a[right]]++;
  int ans_right = -1, ans_left = 0;
  while (left <= n) {
    right++;
    while (right <= n) {
      already[a[right]]++;
      if (already[a[right]] == 1) counter++;
      if (counter > k) break;
      right++;
    }
    if (counter > k) {
      already[a[right]]--;
      right--;
      counter--;
      if (right - left + 1 > ans_right - ans_left + 1) {
        ans_right = right;
        ans_left = left;
      }
      for (left = left; left <= right; left++) {
        already[a[left]]--;
        if (already[a[left]] == 0) {
          counter--;
          break;
        }
      }
    } else if (right > n) {
      if (right - left > ans_right - ans_left + 1) {
        ans_right = right - 1;
        ans_left = left;
      }
    }
    left++;
  }
  printf("%d %d", ans_left, ans_right);
  return 0;
}