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 x[1000006];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  int a[500005], n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int c = 0, d = -1, cnt = 0;
  int i, j, p = -1;
  while (c < n) {
    x[a[c]]++;
    if (x[a[c]] == 1) cnt++;
    if (cnt <= k) {
      if (c - d > p) {
        p = c - d;
        i = c + 1;
        j = d + 2;
      }
    }
    if (cnt > k) {
      while (cnt > k) {
        d++;
        x[a[d]]--;
        if (x[a[d]] == 0) cnt--;
      }
    }
    c++;
  }
  cout << j << " " << i << endl;
}
