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;
const double pi = 2 * acos(0);
const long long N = 1e9 + 5;
long long i, j, t;
int w[1000015];
int main() {
  long long n, b = 0, e = 0, r = 0, p = 0, s;
  cin >> n >> s;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  i = 0;
  j = 0;
  while (j <= n - 1 && i <= j) {
    while (j <= n - 1 && p <= s) {
      if (w[v[j]] == 0 && s == p) {
        break;
      }
      if (w[v[j]] == 0) {
        p++;
      }
      w[v[j]]++;
      j++;
    }
    if (j - i > r) {
      r = j - i;
      b = i;
      e = j - 1;
    }
    w[v[i]]--;
    if (w[v[i]] == 0) {
      p--;
    }
    i++;
  }
  cout << b + 1 << " " << e + 1;
}
