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>
int n, k, countx, d[500010], hashx[1000010];
void in(int p) {
  if (!hashx[d[p]]) countx++;
  hashx[d[p]]++;
}
void out(int p) {
  hashx[d[p]]--;
  if (!hashx[d[p]]) countx--;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &d[i]);
  int p = 0, q = -1, l = 0, r = -1;
  while (q + 1 < n) {
    in(++q);
    while (p < q && countx > k) out(p++);
    if (countx <= k && q - p + 1 > r - l + 1) l = p + 1, r = q + 1;
  }
  printf("%d %d", l, r);
}
