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 c[1000006];
int main() {
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  int a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int max_len = -1, len = 0, l = 1, r = 1, pt1 = 1, pt2 = 1, dist_val = 0;
  for (int i = 1; i <= n; i++) {
    if (c[a[i]] == 0) {
      if (dist_val < k) {
        dist_val++;
        len++;
        pt2 = i;
        c[a[i]] = 1;
        if (max_len < len) {
          max_len = len;
          l = pt1;
          r = pt2;
        }
      } else {
        c[a[i]] = 1;
        len++;
        pt2 = i;
        while (c[a[pt1]] != 1) {
          c[a[pt1]]--;
          pt1++;
        }
        c[a[pt1]] = 0;
        pt1++;
        len = pt2 - pt1 + 1;
        if (max_len < len) {
          max_len = len;
          l = pt1;
          r = pt2;
        }
      }
    } else {
      c[a[i]]++;
      pt2 = i;
      len++;
      if (max_len < len) {
        max_len = len;
        l = pt1;
        r = pt2;
      }
    }
  }
  cout << l << " " << r;
}
