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);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  map<int, int> s;
  int best = -1;
  int left = -1, right = -1;
  int pos = 0;
  for (int i = 0; i < n; ++i) {
    if (s.find(a[i]) != s.end()) {
      s[a[i]] = s[a[i]] + 1;
    } else {
      s[a[i]] = 1;
    }
    int d = s.size();
    while (d > k) {
      if (s[a[pos]] > 1) {
        s[a[pos]] -= 1;
      } else {
        s.erase(s.find(a[pos]));
      }
      d = s.size();
      ++pos;
    }
    if (i - pos > best) {
      left = pos;
      right = i;
      best = i - pos;
    }
  }
  cout << left + 1 << ' ' << right + 1 << endl;
  return 0;
}
