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;
int uniq = 0;
int c[1000005];
int a[1000005];
int main() {
  int n, k, i, j = 0, x, y;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  i = 0;
  j = 0;
  c[a[0]]++;
  uniq = 1;
  int len = 0, ax, ay;
  while (i < n && j < n) {
    if (uniq <= k) {
      j++;
      x = a[j];
      if (c[x] == 0) uniq++;
      c[x]++;
    } else {
      x = a[i++];
      c[x]--;
      if (c[x] == 0) uniq--;
    }
    if ((j - i) > len) {
      ax = j, ay = i + 1;
      len = j - i;
    }
  }
  cout << ay << " " << ax << endl;
}
