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 N = 3e5 + 10;
struct Segment {
  int left, right;
} seg[N / 2 + 5];
map<int, bool> m;
int a[N];
int n, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  m.clear();
  ans = 1;
  seg[ans].left = 1;
  m[a[1]] = true;
  int i = 2;
  while (i <= n) {
    if (m[a[i]]) {
      seg[ans].right = i;
      if (i == n) break;
      ans++;
      i++;
      seg[ans].left = i;
      m.clear();
    }
    m[a[i]] = true;
    i++;
  }
  if (i > n) {
    if (ans == 1) {
      printf("-1\n");
      return 0;
    }
    ans--;
    seg[ans].right = n;
  }
  printf("%d\n", ans);
  for (int i = 1; i <= ans; i++) printf("%d %d\n", seg[i].left, seg[i].right);
  return 0;
}