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>
int a[500000];
int cnt[1000001];
int set_size = 0;
inline void add(int e) {
  cnt[e]++;
  if (cnt[e] == 1) set_size++;
}
inline void remove(int e) {
  cnt[e]--;
  if (cnt[e] == 0) set_size--;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
  }
  int j = 0;
  int ansl = 0;
  int ansr = 0;
  add(a[0]);
  for (int i = 0; i < n; ++i) {
    if (i > 0) {
      remove(a[i - 1]);
    }
    if (j < i) {
      j = i;
      add(a[j]);
    }
    while (j + 1 < n) {
      if (set_size == k && cnt[a[j + 1]] != 0) {
        add(a[++j]);
      } else if (set_size < k) {
        add(a[++j]);
      } else
        break;
    }
    if (j - i > ansr - ansl) {
      ansl = i;
      ansr = j;
    }
  }
  std::cout << (ansl + 1) << ' ' << (ansr + 1) << '\n';
  return 0;
}