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 = 5e5 + 5;
int d, n, m, k;
int l, r, ans;
int a[MAXN];
map<int, int> mp;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int i = 1, j = 1;
  while (j <= n) {
    if (mp.size() <= k)
      mp[a[j++]]++;
    else
      while (mp.size() > k) {
        mp[a[i]]--;
        if (mp[a[i]] <= 0) mp.erase(a[i]);
        i++;
      }
    if (mp.size() <= k && ans < j - i) {
      ans = j - i;
      l = i;
      r = j;
    }
  }
  cout << l << " " << r - 1;
}
