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 N = 1e6 + 9;
void get_input(int &n, int &k, vector<int> &num) {
  cin >> n >> k;
  num.resize(n);
  for (int i = 0; i < n; ++i) cin >> num[i];
}
int main() {
  int n, k;
  vector<int> num;
  get_input(n, k, num);
  vector<int> cnt(N, 0);
  cnt[num[0]]++;
  int counter = 1, ans = 1;
  int ans_l = 0, ans_r = 0;
  int r = 0;
  for (int l = 0; l < n; ++l) {
    while (r < (n - 1) && (counter < k || cnt[num[r + 1]] != 0)) {
      int try_add = num[r + 1];
      if (cnt[try_add] == 0) counter++;
      cnt[try_add]++;
      r++;
    }
    if (r - l + 1 > ans) {
      ans = r - l + 1;
      ans_l = l;
      ans_r = r;
    }
    int try_del = num[l];
    cnt[try_del]--;
    if (cnt[try_del] == 0) counter--;
  }
  cout << ans_l + 1 << ' ' << ans_r + 1;
  return 0;
}