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() {
  std::ios_base::sync_with_stdio(0);
  std::cin.tie(0);
  std::cout.tie(0);
  int k, n;
  cin >> n >> k;
  int a[n];
  int ans = 0;
  map<int, int> ct;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> a[i];
  }
  int l = 0, r = 0;
  int j = 0;
  ct.clear();
  for (long long i = 0; i < (long long)n; i++) {
    while (j < n) {
      if (++ct[a[j]] == 1) ans++;
      if (ans > k) {
        if (--ct[a[j]] == 0) ans--;
        break;
      }
      j++;
    }
    if (r - l < j - i) {
      r = j;
      l = i;
    }
    if (--ct[a[i]] == 0) ans--;
  }
  cout << l + 1 << " " << r << "\n";
  return 0;
}
