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() {
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<pair<int, int> > v;
  map<int, int> m;
  int pre = 1;
  int i = 1;
  for (; i <= n; i++) {
    m[a[i]]++;
    if (m[a[i]] == 2) {
      m.clear();
      v.push_back(make_pair(pre, i));
      pre = i + 1;
    }
  }
  if (v.size() == 0) {
    cout << "-1" << endl;
    return 0;
  }
  if (i == n + 1) {
    if (v[v.size() - 1].second != n) {
      v[v.size() - 1].second = n;
    }
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << endl;
  return 0;
}