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;
map<int, int> cnt;
int where_now, times, ans, start, last, first;
deque<int> check;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k, x;
  cin >> n >> k;
  int inp[n];
  for (int i = 0; i < n; i++) {
    cin >> inp[i];
    cnt.insert(pair<int, int>(inp[i], 0));
  }
  while (where_now != n) {
    while (times == k) {
      start++;
      x = check.front();
      cnt[x]--;
      check.pop_front();
      if (cnt[x] == 0) times--;
    }
    while (times != k && where_now != n) {
      if (cnt[inp[where_now]] == 0) {
        times++;
        cnt[inp[where_now]]++;
        check.push_back(inp[where_now]);
        where_now++;
      }
      while (cnt[inp[where_now]] != 0) {
        cnt[inp[where_now]]++;
        check.push_back(inp[where_now]);
        where_now++;
      }
    }
    int x = check.size();
    if (x > ans) {
      ans = x;
      first = start;
    }
  }
  cout << first + 1 << " " << ans + first;
  return 0;
}
