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;
  scanf("%d", &n);
  ;
  int arr[n + 1], l, r;
  for (int x = 1; x <= n; x++) scanf("%d", &arr[x]);
  ;
  vector<pair<int, int>> res;
  map<int, int> freq;
  l = 1;
  for (int x = 1; x <= n; x++) {
    freq[arr[x]]++;
    if (freq[arr[x]] == 2) {
      res.emplace_back(make_pair(l, x));
      l = x + 1;
      freq.clear();
    }
  }
  if (res.size() == 0) return 0 * printf("-1");
  printf("%d\n", (int)res.size());
  for (int x = 0; x < res.size() - 1; x++)
    printf("%d %d\n", res[x].first, res[x].second);
  printf("%d %d", res[res.size() - 1].first, max(n, res[res.size() - 1].first));
}
