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 freq[1000005] = {};
int n, k;
vector<int> number;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int num;
    cin >> num;
    number.push_back(num);
  }
  int def = 0, sumofNumebr = 0;
  int l1 = 0, r1 = 0;
  for (int i = 0, j = 0; i < n; i++) {
    while (sumofNumebr > k && j < i) {
      freq[number[j]]--;
      if (freq[number[j]] == 0) sumofNumebr--;
      j++;
    }
    freq[number[i]]++;
    if (freq[number[i]] == 1) sumofNumebr++;
    if (sumofNumebr <= k && (i - j > def)) {
      l1 = j, r1 = i, def = i - j;
    }
  }
  cout << l1 + 1 << " " << r1 + 1 << endl;
  return 0;
}
