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;
int main() {
  int n, good_segments = 0, i = 1, count = 0, a_i = 0;
  map<long long, int> pearl_type;
  vector<pair<int, int> > idx;
  scanf("%d", &n);
  int nTmp = n;
  while (n > 0) {
    scanf("%d", &a_i);
    pearl_type[a_i]++;
    count++;
    n--;
    if (pearl_type[a_i] == 2) {
      good_segments++;
      pearl_type.clear();
      idx.push_back(make_pair(i, i + count - 1));
      i = i + count;
      count = 0;
    }
  }
  int t = nTmp - (i - 1);
  if ((nTmp - (i - 1)) != 0 && idx.size() > 0) {
    idx[idx.size() - 1].second += t;
  }
  if (good_segments <= 0) {
    printf("%d\n", -1);
  } else {
    printf("%d\n", good_segments);
    for (unsigned int i = 0; i < idx.size(); i++) {
      printf("%d %d\n", idx[i].first, idx[i].second);
    }
  }
  return 0;
}
