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 v[1000001 / 2 + 1], cate_folosite[1000001];
int citeste_rapid_int_pozitiv() {
  char c;
  int r = 0;
  while ((c = cin.get()) && c >= '0' && c <= '9') r = (r * 10) + (c - '0');
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n = citeste_rapid_int_pozitiv(), k = citeste_rapid_int_pozitiv();
  int i;
  for (i = 1; i <= n; ++i) v[i] = citeste_rapid_int_pozitiv();
  int st = 1, dr = 1, st_ales = 1, dr_ales = 1;
  int cnt = 1;
  cate_folosite[v[dr]] = 1;
  while (st <= n && dr <= n) {
    if (cnt <= k) {
      if (dr - st + 1 > dr_ales - st_ales + 1) dr_ales = dr, st_ales = st;
      ++dr;
      if (cate_folosite[v[dr]] == 0)
        ++cnt, ++cate_folosite[v[dr]];
      else
        ++cate_folosite[v[dr]];
    } else {
      --cate_folosite[v[st]];
      if (cate_folosite[v[st]] == 0) --cnt;
      ++st;
    }
  }
  cout << st_ales << ' ' << dr_ales;
  return 0;
}
