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;
template <typename T>
using V = vector<T>;
const double Pi = acos(-1.0);
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<int> xs(n);
  for (int i = 0; i < int(n); i++) cin >> xs[i];
  int left = 0, best = 0;
  int curleft = 0, curlen = 0;
  map<int, int> mp;
  for (int i = 0; i < int(n); i++) {
    int a = xs[i];
    mp[a]++;
    curlen++;
    while (mp.size() > k) {
      if (--mp[xs[curleft]] == 0) {
        mp.erase(xs[curleft]);
      }
      curlen--;
      curleft++;
    }
    if (curlen > best) {
      best = curlen;
      left = curleft;
    }
  }
  cout << left + 1 << ' ' << left + best << endl;
}
