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 a1[1000001];
int main(void) {
  int n, k;
  cin >> n >> k;
  int *a = new int[n];
  int c = 0, p = 0, l = 0, r = -55, c2 = 0, mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n;) {
    if (a1[a[i]] == 0 && c == k) {
      if (c2 > mx) {
        mx = c2;
        l = p;
        r = i - 1;
      }
      a1[a[p]]--;
      if (a1[a[p]] == 0) {
        c--;
      }
      p++;
      c2--;
    } else {
      c2++;
      a1[a[i]]++;
      if (a1[a[i]] == 1) c++;
      i++;
    }
  }
  if (c2 > mx) {
    r = n - 1;
    if (l + 1 != 0) l = p;
  }
  cout << (l + 1) << " " << (r + 1);
}
