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 MOD = 1e9 + 7;
int Scan() {
  int res = 0, ch, flag = 0;
  if ((ch = getchar()) == '-')
    flag = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return flag ? -res : res;
}
int a[500000 + 5];
int cnt[1000000 + 5];
int main() {
  int n, k;
  scanf("%d%d", &(n), &(k));
  for (int i = 0; i < n; i++) scanf("%d", &(a[i]));
  int last = 0;
  int ansl = 0, ansr = 0;
  int num = 0;
  memset(cnt, 0, sizeof cnt);
  for (int i = 0; i < n; i++) {
    for (int j = last; j < n; j++) {
      if (cnt[a[j]] == 0) num++;
      if (num > k) {
        last = j;
        if (j - i - 1 > ansr - ansl) {
          ansr = j - 1;
          ansl = i;
        }
        if (cnt[a[j]] == 0) num--;
        break;
      }
      if (j == n - 1) {
        last = j;
        if (j - i > ansr - ansl) {
          ansr = j;
          ansl = i;
        }
      }
      cnt[a[j]]++;
    }
    cnt[a[i]]--;
    if (cnt[a[i]] == 0) num--;
    if (last == n - 1) break;
  }
  printf("%d %d\n", ansl + 1, ansr + 1);
  return 0;
}