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 a[500009], n, k, i, cnt, p, has[1000009], ansl, ansr, ma = 0;
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  p = 1, cnt = 0;
  for (i = 1; i <= n & cnt < k; i++) {
    if (has[a[i]] == 0) cnt++;
    has[a[i]]++;
  }
  ma = i - p - 1;
  ansl = 1;
  ansr = i - 1;
  i--;
  for (; i <= n;) {
    if (cnt == k + 1) {
      has[a[p]]--;
      if (has[a[p]] == 0) {
        cnt--;
      }
      p++;
      if (i - p > ma && cnt == k) {
        ma = i - p;
        ansl = p, ansr = i;
      }
    } else {
      i++;
      if (has[a[i]] == 0) {
        cnt++;
      }
      has[a[i]]++;
      if (i - p > ma && cnt == k) {
        ma = i - p;
        ansl = p, ansr = i;
      }
    }
    if (i - p > ma && cnt == k) {
      ma = i - p;
      ansl = p, ansr = i;
    }
  }
  cout << ansl << " " << min(ansr, n);
}
