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;
using namespace std;
int arr[500011];
vector<long long> v;
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k, cnt = 0, start = 0, a = 0, b = 0, end, mx = -1;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    if (cnt < k) {
      if (m[arr[i]] > 0) {
        m[arr[i]]++;
        b++;
      } else {
        m[arr[i]]++;
        cnt++;
        b++;
      }
    } else if (cnt == k && m[arr[i]] > 0) {
      m[arr[i]]++;
      b++;
    } else {
      while (cnt >= k) {
        if (m[arr[a]] > 1) {
          m[arr[a]]--;
          a++;
        } else if (m[arr[a]] == 1) {
          m[arr[a]]--;
          a++;
          cnt--;
        }
      }
      i--;
    }
    if (b - a > mx) {
      start = a;
      end = b;
      mx = b - a;
    }
  }
  cout << start + 1 << " " << end << endl;
}