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 main(void) {
  int n, k, a, track, end, dau = 0, cuoi = 0, length = 1, max = 0;
  vector<int> list;
  int max_length = 1, diff = 0;
  bool xong = false;
  int j, count = 0, l = 0;
  vector<vector<int>> a_vector;
  cin >> n >> k;
  for (int q = 0; q < n; q++) {
    cin >> a;
    list.push_back(a);
  }
  vector<int> cnt(1000000, 0);
  for (int i = 0; i < n; i++) {
    cnt[list[i]]++;
    if (cnt[list[i]] == 1) diff++;
    while (diff > k) {
      cnt[list[l]]--;
      if (cnt[list[l]] == 0) diff--;
      l++;
    }
    if (max_length < i - l + 1) {
      dau = l;
      cuoi = i;
      max_length = cuoi - dau + 1;
    }
  }
  cout << dau + 1 << " " << cuoi + 1 << endl;
  return 0;
}
