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;
const int maxn = 500005;
map<int, int> vis;
int n, k, a[maxn], ans, tot, now, l, r;
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ans = -1;
  now = 1;
  l = r = 1;
  for (int i = 1; i <= n;) {
    while (tot <= k && now <= n) {
      if (!vis[a[now]]) tot++;
      vis[a[now]]++;
      now++;
      if (now > n && tot <= k) {
        if (now - i > ans) ans = now - i, l = i, r = n;
        break;
      }
    }
    if (now - 1 - i > ans) ans = now - 1 - i, l = i, r = now - 2;
    if (now > n) break;
    while (tot > k && i <= n) {
      vis[a[i]]--;
      if (!vis[a[i]]) tot--;
      i++;
    }
  }
  printf("%d %d\n", l, r);
  return 0;
}
