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;
const int MAX = 1000000;
int N;
int main() {
  cin >> N;
  vector<int> sol;
  set<int> seen;
  for (int i = 0; i < N; ++i) {
    int x;
    cin >> x;
    if (seen.find(x) == seen.end()) {
      seen.insert(x);
    } else {
      seen.clear();
      sol.push_back(i + 1);
    }
  }
  if (sol.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (!seen.empty()) {
    sol[sol.size() - 1] = N;
  }
  cout << sol.size() << endl;
  int prev = 1;
  for (int last : sol) {
    cout << prev << " " << last << endl;
    prev = last + 1;
  }
  return 0;
}
