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 main() {
  ios_base::sync_with_stdio(false);
  int n, m, pos = 0, pos2 = 0, mx = -2, mx_pos, mx_pos2;
  scanf("%d %d", &n, &m);
  if (n == m) {
    cout << 1 << ' ' << n << endl;
    return 0;
  }
  vector<int> a(n);
  set<int> st;
  map<int, int> mp;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  while (pos2 != n) {
    st.insert(a[pos2]);
    mp[a[pos2]]++;
    if (st.size() > m) {
      while (st.size() > m) {
        mp[a[pos]]--;
        if (mp[a[pos]] == 0) st.erase(a[pos]);
        pos++;
      }
    } else if (pos2 - pos > mx)
      mx_pos = pos, mx_pos2 = pos2, mx = pos2 - pos;
    pos2++;
  }
  cout << mx_pos + 1 << ' ' << mx_pos2 + 1 << endl;
}
