Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  map<long long int, long long int> a;
  long long int c = 0, n, i, begin = 1, x, k = 0;
  scanf("%I64d", &n);
  long long int res[300300][2];
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    a[x]++;
    if (a[x] >= 2 && begin != i) {
      c++;
      res[k][0] = begin;
      res[k][1] = i;
      begin = i + 1;
      k++;
      a.clear();
    }
  }
  if (c == 0) {
    printf("-1\n");
    return 0;
  }
  res[k - 1][1] = n;
  printf("%I64d\n", c);
  for (i = 0; i < k; i++) {
    printf("%I64d %I64d\n", res[i][0], res[i][1]);
  }
  return 0;
}