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;
long long mod = 1e9 + 7;
map<int, int> m;
int a[500005];
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  scanf("%d", &(n));
  scanf("%d", &(k));
  int l = 1, lp = 1, rp = 1;
  for (int i = (1); i <= (n); i++) scanf("%d", &(a[i]));
  int ans = 0;
  for (int i = (1); i <= (n); i++) {
    m[a[i]]++;
    if (m.size() > k) {
      if (m[a[l]] == 1)
        m.erase(a[l]);
      else
        m[a[l]]--;
      l++;
    }
    if (i - l + 1 > ans) ans = i - l + 1, lp = l, rp = i;
  }
  cout << lp << " " << rp << endl;
  return 0;
}