Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 100;
int arr[maxn], prec[maxn];
int main() {
  int n, k, l = 0, r = -1, len = 0, cnt = 0, ansl = 1, ansr = 1;
  cin >> n >> k;
  for (int i = 0; i < maxn; i++) prec[i] = -1;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  while (r < n - 1) {
    r++;
    if (prec[arr[r]] == -1) cnt++;
    prec[arr[r]] = r;
    while (cnt > k) {
      if (prec[arr[l]] == l) {
        prec[arr[l]] = -1;
        cnt--;
      }
      l++;
    }
    len = r - l + 1;
    if (len > ansr - ansl + 1) {
      ansl = l + 1;
      ansr = r + 1;
    }
  }
  cout << ansl << " " << ansr << '\n';
}