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;
long long u[1000003] = {0};
long long z[1000003];
int main() {
  long long n, k;
  cin >> n >> k;
  long long x[n + 5];
  for (long long i = 1; i <= n; i++) {
    cin >> x[i];
  }
  long long c = 0;
  long long r = 1;
  long long ans = 0;
  long long a = 0;
  long long b = 0;
  for (long long i = 1; i <= n; i++) {
    u[x[i]]++;
    if (u[x[i]] == 1) {
      c++;
    }
    while (c > k) {
      u[x[r]]--;
      if (u[x[r]] == 0) {
        c--;
      }
      r++;
    }
    if (c == k) {
      long long t = (i - r) + 1;
      if (t > ans) {
        ans = t;
        a = r;
        b = i;
      }
    }
  }
  if (((n - r) + 1) > ans) {
    ans = (n - r);
    a = r;
    b = n;
  }
  cout << a << " " << b << endl;
}
