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 = 5e5 + 5;
int a[N];
int st[N * 10];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int cnt = 0;
  int l = 0, r = 0;
  int dist = 0;
  for (int i = 0, j = 0; i < n; i++) {
    if (!st[a[i]]) {
      st[a[i]]++;
      cnt++;
    } else
      st[a[i]]++;
    while (cnt > k) {
      st[a[j]]--;
      if (!st[a[j]]) cnt--;
      j++;
    }
    if (i - j + 1 > dist) {
      dist = i - j + 1;
      l = j, r = i;
    }
  }
  cout << l + 1 << ' ' << r + 1 << endl;
  return 0;
}