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 n, k, a[500001], cnt[1000001];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]] = 0;
  }
  int l = 1, r = 1;
  cnt[a[r]] = 1;
  for (int i = l, j = r, t = 1; j < n;) {
    for (; t <= k && j <= n;) {
      if (++cnt[a[++j]] == 1) {
        ++t;
      }
    }
    if (r - l < j - i - 1) {
      l = i;
      r = j - 1;
    }
    for (; t > k && i <= n;) {
      if (--cnt[a[i++]] == 0) {
        --t;
      }
    }
  }
  cout << l << ' ' << r << endl;
}
