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;
  scanf("%d %d", &n, &k);
  int A[5 * ((int)1e5)] = {};
  for (int x = 0; x < n; x++) scanf("%d", &A[x]);
  int best = 0, last = 0;
  map<int, int> m;
  m[A[0]]++;
  pair<int, int> p = {1, 1};
  for (int x = 0; x < n; x++) {
    if (x > 0) m[A[x - 1]]--;
    if (x > 0 && m[A[x - 1]] == 0) m.erase(A[x - 1]);
    while (last + 1 < n && (m.count(A[last + 1]) || m.size() < k))
      m[A[last + 1]]++, last++;
    if (last - x + 1 > best) p = {x, last}, best = last - x + 1;
  }
  printf("%d %d\n", p.first + 1, p.second + 1);
}
