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;
long long M = 1e9 + 7;
long long MM = 1e9 + 11;
long long hp = 31;
long long hpc = 53;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<tuple<int, int, int> > v;
  int a[n];
  for (long long i = (0); i < (n); i++) cin >> a[i];
  int i1, i2, j = 0;
  map<int, int> m;
  for (long long i = (0); i < (n); i++) {
    if (i > 0) {
      m[a[i - 1]]--;
      if (m[a[i - 1]] == 0) {
        m.erase(a[i - 1]);
      }
    }
    while (1) {
      if (m.size() > k) {
        i1 = i + 1;
        i2 = j - 1;
        break;
      }
      if (j == n) {
        i1 = i + 1;
        i2 = j;
        break;
      }
      m[a[j]]++;
      j++;
    }
    v.push_back(make_tuple(i2 - i1, i1, i2));
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  cout << get<1>(v[0]) << " " << get<2>(v[0]) << endl;
}
