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, x;
  cin >> n;
  set<int> st;
  int j = 1, m[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (st.empty() || st.find(x) == st.end())
      st.insert(x);
    else
      m[j++] = i, st.clear();
  }
  m[j - 1] = n;
  if (j == 1)
    cout << "-1";
  else {
    cout << j - 1 << endl;
    cout << "1 " << m[1] << endl;
    for (int i = 1; i < j - 1; i++) cout << m[i] + 1 << " " << m[i + 1] << endl;
  }
  return 0;
}
