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>
using namespace std;
const int N = 5e5 + 5;
int n, k, a[N], fr[1234567];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  int diff = 0, l = 1, r = 1, MxDiff = 0, last = -1;
  for (int i = 0, j = 0; j < n;) {
    if (fr[a[j]] == 0) diff++;
    if (last != j) fr[a[j]]++;
    last = j;
    if (diff > k) {
      fr[a[i]]--;
      if (fr[a[i]] == 0) diff--;
      i++;
    } else {
      if (MxDiff < j - i + 1) {
        l = i + 1;
        r = j + 1;
        MxDiff = j - i + 1;
      }
      ++j;
    }
  }
  printf("%d %d", l, r);
  return 0;
}