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 long long MAX = 5e5 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 12;
const long long N = 1002;
set<long long> ss;
map<long long, long long> mm;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cout.tie(NULL);
  ;
  long long tt = 1;
  while (tt--) {
    long long n, k;
    cin >> n >> k;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0, i = 0, j = 0, x = 0, y = 0;
    while (i < n) {
      ss.insert(a[i]);
      mm[a[i]]++;
      if (ss.size() > k) {
        long long k = j;
        while (1) {
          mm[a[k]]--;
          if (mm[a[k]] == 0) {
            ss.erase(a[k]);
            break;
          }
          k++;
        }
        j = k + 1;
      }
      if (ans < i - j + 1) {
        ans = i - j + 1;
        y = i;
        x = j;
      }
      i++;
    }
    cout << x + 1 << " " << y + 1;
  }
}
