Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxx = 5e5 + 20;
int n, k, t, a[maxx];
pair<int, int> ans;
map<int, int> m;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int p = 0;
  ans.first = -1;
  for (int i = 0; i < n; i++) {
    while (p < n && t <= k) {
      if (m.find(a[p]) == m.end()) {
        if (t == k) break;
        t++, m[a[p]] = 1;
      } else
        m[a[p]]++;
      p++;
    }
    if (ans.first == -1 || ans.second - ans.first < p - i)
      ans.first = i, ans.second = p;
    m[a[i]]--;
    if (m[a[i]] == 0) t--, m.erase(a[i]);
  }
  cout << ans.first + 1 << " " << ans.second << endl;
  return 0;
}