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 i, j, n, k = 0;
  scanf("%d", &n);
  int a[n];
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  map<int, bool> my;
  vector<pair<int, int> > v;
  for (i = 0, j = 0; i < n; i++) {
    if (!my[a[i]])
      my[a[i]] = true;
    else {
      v.push_back(make_pair(j + 1, i + 1));
      j = i + 1;
      k++;
      my.clear();
    }
  }
  if (k == 0)
    printf("-1\n");
  else {
    printf("%d\n", k);
    for (i = 0; i < k - 1; i++) printf("%d %d\n", v[i].first, v[i].second);
    printf("%d %d\n", v[k - 1].first, n);
  }
}
