Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  map<int, int> ma;
  set<int> se;
  int n;
  cin >> n;
  int l, data;
  l = 1;
  int count = 0;
  for (int i = 1; i <= n; i++) {
    cin >> data;
    if (se.find(data) != se.end()) {
      ma[l] = i;
      se.clear();
      l = i + 1;
      count++;
    } else {
      se.insert(data);
    }
  }
  if (count == 0) {
    cout << -1;
  } else {
    map<int, int>::iterator iter;
    iter = ma.end();
    iter--;
    iter->second = n;
    cout << count << endl;
    for (iter = ma.begin(); iter != ma.end(); iter++) {
      cout << iter->first << " " << iter->second << endl;
    }
  }
  return 0;
}