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;
long long arr[5000005];
int main() {
  long long n, k;
  scanf("%lld", &n);
  ;
  scanf("%lld", &k);
  ;
  for (long long i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  ;
  long long l, r, mx, a, b;
  mx = 0;
  l = r = 1;
  map<long long, long long> m;
  map<long long, long long>::iterator it;
  while (r <= n) {
    if (m.size() <= k) {
      if ((r - l) > mx) {
        mx = r - l;
        a = l;
        b = r - 1;
      }
      it = m.find(arr[r]);
      if (it == m.end()) {
        m[arr[r]] = 1;
      } else {
        m[arr[r]]++;
      }
      r++;
    } else {
      it = m.find(arr[l]);
      if ((*it).second == 1) {
        m.erase(it);
      } else {
        m[arr[l]]--;
      }
      l++;
    }
  }
  if (m.size() <= k) {
    if ((r - l) > mx) {
      mx = r - l;
      a = l;
      b = r - 1;
    }
  }
  printf("%lld %lld", a, b);
  return 0;
}
