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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> mp;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = a[i] + 1;
  }
  long long dif = 0;
  long long l = 0, r = 0;
  vector<pair<long long, long long> > vect;
  while (l <= r and r < n) {
    if (mp[a[r]] == 0) {
      dif++;
    }
    mp[a[r]]++;
    if (k >= dif) {
      vect.push_back({l, r});
    } else if (k < dif) {
      while (l <= r and dif > k) {
        if (mp[a[l]] == 1) {
          dif--;
        }
        mp[a[l]]--;
        l++;
        if (k >= dif) {
          vect.push_back({l, r});
          break;
        }
      }
    }
    r++;
  }
  if (dif <= k) {
    vect.push_back({l, r - 1});
  }
  long long ans1, ans2;
  long long m = 0;
  for (long long i = 0; i < vect.size(); i++) {
    if (vect[i].first <= vect[i].second) {
      if (vect[i].second - vect[i].first + 1 > m) {
        m = vect[i].second - vect[i].first + 1;
        ans1 = vect[i].first;
        ans2 = vect[i].second;
      }
    }
  }
  ans1++;
  ans2++;
  cout << ans1 << " " << ans2 << '\n';
  return 0;
}