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 const MAX = 3e5 + 5;
int n, a[MAX], l[MAX], r[MAX], cnt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  set<int> st;
  cnt = 0;
  l[cnt] = 1;
  for (int i = 1; i <= n; i++) {
    if (st.find(a[i]) == st.end())
      st.insert(a[i]);
    else {
      r[cnt++] = i;
      st.clear();
      l[cnt] = i + 1;
    }
  }
  if (cnt == 0)
    printf("-1\n");
  else {
    if (r[cnt] == 0) r[cnt - 1] = n;
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; i++)
      if (r[i]) printf("%d %d\n", l[i], r[i]);
  }
}