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, k, i, j, l, r, cnt, maxx = -1;
  scanf("%d%d", &n, &k);
  int a[n + 1];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  map<int, int> m;
  map<int, int>::iterator it;
  for (i = 0, j = 0; i < n; i++) {
    m[a[i]]++;
    if (m.size() > k) {
      while (m.size() > k) {
        it = m.find(a[j]);
        j++;
        if (it->second > 1) {
          --it->second;
        } else {
          m.erase(it);
        }
      }
    }
    cnt = i - j + 1;
    if (cnt > maxx) {
      maxx = cnt;
      l = j;
      r = i;
    }
  }
  printf("%d %d", ++l, ++r);
  return 0;
}
