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_n = 300000;
int N;
int type[max_n];
vector<pair<int, int> > ans;
bool solve() {
  int start = 0;
  set<int> exist;
  for (int i = 0; i < N; i++) {
    if (exist.count(type[i]) == 0) {
      exist.insert(type[i]);
    } else {
      ans.push_back(pair<int, int>(start, i));
      start = i + 1;
      exist.clear();
    }
  }
  if (exist.size() != 0) {
    if (ans.size() == 0) return false;
    pair<int, int> back = ans.back();
    ans.pop_back();
    ans.push_back(pair<int, int>(back.first, N - 1));
  }
  return true;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &type[i]);
  if (solve() == false)
    puts("-1");
  else {
    printf("%d\n", int(ans.size()));
    for (auto p : ans) {
      printf("%d %d\n", p.first + 1, p.second + 1);
    }
  }
  return 0;
}
