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 maxSize = 1e6 + 10;
int repetition[maxSize];
int main() {
  int n, k, i;
  cin >> n >> k;
  int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  int ans1 = 0, ans2 = 0, l = 0;
  int distincts = 0;
  for (i = 0; i < n; i++) {
    repetition[a[i]]++;
    if (repetition[a[i]] == 1) distincts++;
    while (distincts > k) {
      repetition[a[l]]--;
      if (repetition[a[l]] == 0) distincts--;
      l++;
    }
    if (ans2 - ans1 < i - l) ans1 = l, ans2 = i;
  }
  cout << ans1 + 1 << " " << ans2 + 1;
}