Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, k, a[500500], mx = -1, l, r;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  map<int, int> mp;
  int i = 0, j = 0, c = 0;
  while (j < n) {
    if (mp[a[j]] == 0) ++c;
    mp[a[j]]++;
    if (c > k) {
      while (c > k && i <= j) {
        mp[a[i]]--;
        if (mp[a[i]] == 0) --c;
        ++i;
      }
    }
    if (j - i > mx) {
      l = i + 1, r = j + 1;
      mx = j - i;
    }
    ++j;
  }
  printf("%d %d", l, r);
  return 0;
}