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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int cnt[1000001];
  memset(cnt, 0, sizeof(cnt));
  long long int j = 0, dis = 0;
  long long int ansl = 0, ansr = 0;
  for (long long int i = 0; i < n; i++) {
    while (j < n) {
      dis += (++cnt[a[j]] == 1);
      if (dis > k) {
        dis -= (--cnt[a[j]] == 0);
        break;
      }
      j++;
    }
    if (ansr - ansl + 1 < j - i + 1) {
      ansl = i;
      ansr = j - 1;
    }
    dis -= (--cnt[a[i]] == 0);
  }
  cout << ansl + 1 << " " << ansr + 1;
}
