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 int maxn = 500005;
int a[maxn];
map<int, int> M;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int r = 1;
  int ansl = 1, ansr = 1;
  for (int i = 1; i <= n; i++) {
    while (r <= n && ((int)M.size() < k ||
                      (int)M.size() == k && M.find(a[r]) != M.end())) {
      M[a[r]]++;
      r++;
    }
    if (r - 1 - i > ansr - ansl) {
      ansl = i;
      ansr = r - 1;
    }
    M[a[i]]--;
    if (M[a[i]] == 0) M.erase(a[i]);
  }
  printf("%d %d", ansl, ansr);
  return 0;
}
