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;
int32_t main() {
  long long t = 1;
  while (t--) {
    long long n, k;
    scanf("%lld%lld", &n, &k);
    long long a[n];
    for (long long i = 0; i < n; i++) scanf("%lld", &a[i]);
    long long left = 0, right = -1, al = 0, ar = 0;
    map<long long, long long> m;
    while (right < n - 1) {
      m[a[++right]]++;
      while (m.size() > k) {
        m[a[left]]--;
        if (m[a[left]] == 0) long long ul = m.erase(a[left]);
        left++;
      }
      if (m.size() <= k && (right - left) >= (ar - al))
        ar = right + 1, al = left + 1;
    }
    printf("%lld %lld", al, ar);
  }
  return 0;
}
