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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int k, n;
  cin >> n >> k;
  int ara[n];
  for (int i = 0; i < n; i++) cin >> ara[i];
  map<int, int> mp;
  int i = 0, j = 0, dis = 0;
  int maxd = -1, ii = 0, jj = 0;
  while (i < n && j < n) {
    int a = ara[j];
    if (mp[a] == 0) {
      dis++;
      mp[a]++;
    } else
      mp[a]++;
    while (dis > k) {
      int b = ara[i];
      if (mp[b] == 1) {
        dis--;
        mp[b]--;
      } else
        mp[b]--;
      i++;
    }
    if (j - i > maxd) {
      maxd = j - i;
      ii = i;
      jj = j;
    }
    j++;
  }
  cout << ii + 1 << " " << jj + 1 << "\n";
}
