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;
map<long long int, int> my;
long long int arr[400000];
int main() {
  int n, i, l, r, cnt;
  scanf("%d", &n);
  pair<int, int> p[n + 10];
  for (i = 1; i <= n; i++) {
    scanf("%i64d", &arr[i]);
  }
  l = 1;
  r = 1;
  cnt = 1;
  bool f = 0;
  for (i = 1; i <= n; i++) {
    f = 0;
    if (my[arr[i]] == 0) {
      my[arr[i]] = i;
    } else {
      p[cnt].first = l;
      p[cnt].second = i;
      l = i + 1;
      cnt++;
      my.clear();
      f = 1;
    }
  }
  if (f == 0) {
    p[cnt - 1].second = n;
  }
  if (cnt == 1) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", cnt - 1);
  for (i = 1; i < cnt; i++) {
    printf("%d %d\n", p[i].first, p[i].second);
  }
}