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[500005];
int main() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  set<pair<int, int> > st;
  map<int, int> mp;
  set<pair<int, int> >::iterator iter;
  int mx = -1;
  int size = 0;
  pair<int, int> mxsize;
  int strt = 1;
  for (int i = 0; i < n; i++) {
    if (mp.find(a[i]) != mp.end()) {
      st.erase(make_pair(mp[a[i]], a[i]));
    }
    mp[a[i]] = i;
    st.insert(make_pair(mp[a[i]], a[i]));
    size++;
    if (st.size() > k) {
      iter = st.begin();
      size = i - iter->first;
      strt = iter->first + 2;
      mp.erase(iter->second);
      st.erase(iter);
    }
    if (size > mx) {
      mx = size;
      mxsize.first = strt;
      mxsize.second = i + 1;
    }
  }
  printf("%d %d\n", mxsize.first, mxsize.second);
}
