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 cont[1000050];
int num[600000];
int main() {
  int i, j, n, k, t, l, r, max;
  for (i = 0; i <= 500000; i++) cont[i] = 0;
  set<int> s;
  t = 1;
  max = 0;
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &num[i]);
  i = 1;
  j = 2;
  l = 1;
  r = 1;
  s.insert(num[1]);
  cont[num[1]] = 1;
  while (j <= n) {
    if (s.count(num[j]) == 0) {
      t++;
      s.insert(num[j]);
    }
    cont[num[j]]++;
    if (t <= k) {
      if (j - i > max) {
        max = j - i;
        l = i;
        r = j;
      }
    } else {
      while (t > k) {
        cont[num[i]]--;
        if (cont[num[i]] == 0) {
          s.erase(num[i]);
          t--;
        }
        i++;
      }
    }
    j++;
  }
  printf("%d %d\n", l, r);
}
