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(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  unordered_map<long long, long long> mp;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  if (mp.size() <= k) {
    cout << 1 << " " << n;
    return 0;
  }
  mp.clear();
  long long m = 0, mi = 0, mj = 0, l = 0, r = 0;
  while (r < n) {
    mp[v[r]]++;
    if (mp.size() > k) {
      if (m < (r - l)) {
        m = r - l;
        mj = r;
        mi = l + 1;
      }
      while (mp.size() > k) {
        mp[v[l]]--;
        if (mp[v[l]] == 0) {
          mp.erase(v[l]);
        }
        l++;
      }
    }
    r++;
  }
  if (m < (r - l)) {
    mi = l + 1;
    mj = r;
  }
  cout << mi << " " << mj;
}
