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
#include <bits/stdc++.h>
using namespace std;
map<int, int> m;
int answer[300000][2];
int arr[300000];
int main() {
  int n, i, temp;
  int cnt = 0;
  scanf("%d", &n);
  map<int, int>::iterator idx;
  int tmp = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &temp);
    idx = m.find(temp);
    if (idx != m.end()) {
      answer[cnt][0] = tmp + 1;
      answer[cnt++][1] = i;
      tmp = i;
      m.clear();
    } else
      m.insert(pair<int, int>(temp, i));
  }
  if (cnt == 0)
    printf("-1\n");
  else {
    answer[cnt - 1][1] = n - 1;
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; i++)
      printf("%d %d\n", answer[i][0] + 1, answer[i][1] + 1);
  }
}
