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;
void faster() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int N = 5e5;
int n, a[N], cnt, k, ans_l, ans_r, l, r;
map<int, int> mp;
int main() {
  faster();
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  ans_l = 1;
  while (r != n) {
    if (!mp[a[r]] && cnt < k)
      cnt++;
    else if (!mp[a[r]]) {
      while (cnt == k) {
        if (mp[a[l]] == 1) cnt--;
        mp[a[l++]]--;
      }
      cnt++;
    }
    mp[a[r]]++;
    if (ans_r - ans_l + 1 < r - l + 1) ans_l = l, ans_r = r;
    r++;
  }
  cout << ans_l + 1 << " " << ans_r + 1 << endl;
}
