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;
bool comp(int a, int b) { return (a > b); }
int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int m[1000001];
int main(void) {
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  int i, a[n], j = 0, l, r, c = 0, ma = 0;
  for (i = 0; i < n; ++i) cin >> a[i];
  for (i = 0; i < n; ++i) {
    if (!m[a[i]]) c++;
    m[a[i]]++;
    if (c > k) {
      if (ma < (i - j)) {
        ma = i - j;
        l = j + 1;
        r = i;
      }
      while (c > k) {
        m[a[j]]--;
        if (m[a[j]] == 0) c--;
        j++;
      }
    }
  }
  if (ma < i - j) {
    l = j + 1;
    r = i;
  }
  cout << l << " " << r;
  return 0;
}