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;
long long read() {
  long long a = 0, b = getchar(), c = 1;
  while (!isdigit(b)) c = b == '-' ? -1 : 1, b = getchar();
  while (isdigit(b)) a = a * 10 + b - '0', b = getchar();
  return a * c;
}
int n, k, ans, sum, res, ret, x, a[500005], b[1000005];
int main() {
  n = read(), k = read();
  for (int i = 0; i < n; i++) a[i] = read();
  for (int i = 0; i < n; i++) {
    if (i != 0) b[a[i - 1]]--, sum -= b[a[i - 1]] == 0;
    while (sum <= k and x < n) {
      if (sum == k and b[a[x]] == 0) break;
      sum += b[a[x]] == 0, b[a[x]]++, x++;
    }
    if (x - i > ans) ans = x - i, res = i + 1, ret = x;
  }
  printf("%d %d", res, ret);
  return 0;
}
