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;
int n, k, cnt, a[555555], flag[2 * 555555];
int main() {
  while (~scanf("%d%d", &n, &k)) {
    memset(flag, 0, sizeof(flag));
    cnt = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (i <= k) {
        if (!flag[a[i]])
          cnt++, flag[a[i]] = 1;
        else
          flag[a[i]]++;
      }
    }
    int l = 1, r = k, ansl = 1, ansr = k;
    while (r < n) {
      if (flag[a[r + 1]] || cnt < k) {
        r++;
        if (flag[a[r]] == 0) cnt++;
        flag[a[r]]++;
        if (ansr - ansl < r - l) ansl = l, ansr = r;
      } else {
        if (flag[a[l]] == 1) cnt--;
        flag[a[l]]--;
        l++;
      }
    }
    printf("%d %d\n", ansl, ansr);
  }
  return 0;
}