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 f[1000000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, d;
  cin >> n >> d;
  vector<int> a(n);
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  };
  set<int> b;
  int l = 0, r = 0, ml = 0, mr = 0, length = -1;
  f[a[0]]++;
  while (l < n) {
    while (r < n && b.size() <= d) {
      b.insert(a[r]);
      if (r - l > length) {
        ml = l + 1;
        mr = r + 1;
        length = r - l;
      }
      r++;
      b.insert(a[r]);
      if (r < n) f[a[r]]++;
    }
    if (f[a[l]] == 1) {
      b.erase(a[l]);
      f[a[l]] = 0;
    } else {
      f[a[l]]--;
    }
    l++;
  }
  cout << ml << " " << mr;
}
