Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int v[500005];
int got[1000005];
int main(void) {
  int n, k;
  scanf(" %d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf(" %d", &v[i]);
  }
  for (int i = 0; i < 1000005; i++) {
    got[i] = 0;
  }
  int l = 0;
  int r = -1;
  int bl = 0;
  int br = 0;
  int big = 0;
  int in = 0;
  while (l < n) {
    if (r < l) {
      r++;
      int x = v[r];
      if (got[x] == 0) {
        in++;
      }
      got[x]++;
    } else {
      if (in <= k && r < n - 1) {
        r++;
        int x = v[r];
        if (got[x] == 0) {
          in++;
        }
        got[x]++;
      } else {
        int x = v[l];
        if (got[x] == 1) {
          in--;
        }
        got[x]--;
        l++;
      }
    }
    if (in <= k && r - l + 1 > big) {
      big = r - l + 1;
      br = r;
      bl = l;
    }
  }
  printf("%d %d\n", bl + 1, br + 1);
  return 0;
}