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() {
  int x, y;
  cin >> x >> y;
  vector<int> vec(x);
  for (int &i : vec) cin >> i;
  map<int, int> mp;
  int sum = 0, cnt = 0, l = 0, r = 0, ll = 0, rr = 0;
  for (int i = 0; i < x; i++) {
    mp[vec[i]]++;
    if (mp.size() <= y)
      l++;
    else {
      if (l - r > ll - rr) {
        rr = r;
        ll = l;
      }
      for (; mp[vec[r]] > 1; r++) {
        mp[vec[r]]--;
      }
      mp.erase(vec[r]);
      if (r > l)
        r--;
      else if (l > r)
        r++;
      l++;
    }
  }
  if (l - r > ll - rr) {
    rr = r;
    ll = l;
  }
  cout << rr + 1 << " " << ll << endl;
  cout << endl;
}
