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;
const double PI = acos(-1);
int a[500000];
int used[1000005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  int start = 1, end = 0, j = 0, diffs = 0;
  for (int i = 0; i < n; ++i) {
    if (diffs != k) {
      for (; j < n; ++j) {
        if (diffs == k && !used[a[j]]) break;
        if (!used[a[j]]) {
          diffs++;
        }
        used[a[j]]++;
      }
    }
    if (end - start < (j - 1) - i) {
      end = j - 1;
      start = i;
    }
    used[a[i]]--;
    if (!used[a[i]]) {
      diffs--;
    }
  }
  printf("%d %d\n", start + 1, end + 1);
  return 0;
}
