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;
const int nsz = 5e5, asz = 1e6;
int n, k, a[nsz + 5], cnt[asz + 5], rl, rr;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int l = 1, r = 1, cur = 0; l <= n; ++l) {
    for (; r <= n && cur < k; cur += !cnt[a[r]]++, ++r)
      ;
    for (; r <= n && cnt[a[r]]; ++cnt[a[r]], ++r)
      ;
    r - l >= rr - rl + 1 && (rl = l, rr = r - 1);
    cur -= !--cnt[a[l]];
  }
  printf("%d %d\n", rl, rr);
}
