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;
template <typename T>
struct myhash : std::hash<T> {
  std::size_t operator()(T const& obj) const {
    auto v = std::hash<T>::operator()(obj);
    v = ((v >> 1) & 0x5555555555555555) | ((v & 0x5555555555555555) << 1);
    return v;
  }
};
int main() {
  unordered_set<int, myhash<int> > used;
  vector<int> ansi, ansj;
  int a, n, i, cur;
  scanf("%d", &n);
  cur = 1;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (used.find(a) != used.end()) {
      ansi.push_back(cur);
      ansj.push_back(i);
      cur = i + 1;
      used.clear();
    } else {
      used.insert(a);
    }
  }
  if (!used.empty() && !ansi.empty()) {
    ansj.back() = n;
  }
  if (ansi.empty())
    printf("-1\n");
  else {
    printf("%d\n", (int)ansi.size());
    for (i = 0; i < (int)ansi.size(); i++) printf("%d %d\n", ansi[i], ansj[i]);
  }
  return 0;
}
