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 a[1234567];
set<int> s;
int pre, ne;
int posl = -1, posr = -1;
int rp[1234567];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int l = 0, r = 0;
  int maxm = 0;
  s.insert(a[0]);
  r = 0;
  rp[a[0]] = 0;
  for (int i = 1; i < n; i++) {
    r++;
    pre = s.size();
    if (s.find(a[i]) == s.end()) s.insert(a[i]);
    ne = s.size();
    rp[a[i]] = i;
    if (ne <= k) {
      if ((r - l + 1) >= maxm) {
        maxm = r - l + 1;
        posl = l;
        posr = r;
      }
    } else {
      l++;
      if (rp[a[l - 1]] < l && s.find(a[l - 1]) != s.end()) s.erase(a[l - 1]);
    }
  }
  int sz = s.size();
  if (sz == k) {
    if ((r - l + 1) >= maxm) {
      maxm = r - l + 1;
      posl = l;
      posr = r;
    }
  }
  printf("%d %d", posl + 1, posr + 1);
  return 0;
}
