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 = 5 * (1e5 + 10);
const int MAX = 1e6 + 10;
const int mod = 1e9 + 7;
const int PI = 3.14159, BASE = 3137;
int n, k;
int arr[MAXN];
int imam[MAX];
int len = 1;
int soll = 1, solr = 1;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  int lo = 0, hi = 0;
  set<int> S;
  while (hi <= n) {
    S.insert(arr[hi]);
    imam[arr[hi]]++;
    while ((int)S.size() > k) {
      imam[arr[lo]]--;
      if (!imam[arr[lo]]) S.erase(arr[lo]);
      lo++;
    }
    if (lo > hi) hi = lo;
    if (hi - lo + 1 > len && (int)S.size() <= k) {
      len = hi - lo + 1;
      soll = lo + 1;
      solr = hi + 1;
    }
    if (hi == n - 1) break;
    hi++;
  }
  printf("%d %d\n", soll, solr);
  return 0;
}
