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() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  map<int, int> mp;
  for (int i = 0; i < n; i++) cin >> a[i];
  int m1 = 0;
  int start = 0, end = 0, x = 0, y = 0;
  while (end < n) {
    mp[a[end]]++;
    if (mp.size() > k) {
      mp[a[start]] -= 1;
      if (mp[a[start]] == 0) mp.erase(a[start]);
      start += 1;
    }
    if (m1 < end - start + 1) {
      m1 = end - start + 1;
      x = start;
      y = end;
    }
    end += 1;
  }
  cout << x + 1 << " " << y + 1;
  return 0;
}
