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 MAXN = 3 * 100 * 1000 + 10;
int arr[MAXN];
vector<int> vec;
set<int> s;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  vec.push_back(0);
  for (int i = 0; i < n; i++) {
    int y = s.size();
    s.insert(arr[i]);
    if (s.size() == y) {
      vec.push_back(i + 1);
      s.clear();
    }
  }
  if (vec.size() == 1) {
    cout << "-1\n";
    return 0;
  }
  if (vec[vec.size() - 1] != n) {
    vec.erase(vec.begin() + vec.size() - 1, vec.begin() + vec.size());
    vec.push_back(n);
  }
  cout << vec.size() - 1 << endl;
  for (int i = 0; i < vec.size() - 1; i++)
    cout << vec[i] + 1 << " " << vec[i + 1] << endl;
}
