Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  while (TESTS--) {
    int i, n, k;
    cin >> n >> k;
    vector<int> v(n);
    map<int, int> mp;
    vector<int> ind1(n + 1), ind2(n + 1);
    for (i = 0; i < n; i++) cin >> v[i];
    int l = 0, sz = 0, r;
    for (r = 0; r < n; r++) {
      mp[v[r]]++;
      while (mp.size() > k) {
        if (--mp[v[l]] == 0) mp.erase(v[l]);
        l++;
      }
      sz = max((r - l + 1), sz);
      if ((r - l + 1) == sz) {
        ind1[sz] = l + 1;
        ind2[sz] = r + 1;
      }
    }
    cout << ind1[sz] << " " << ind2[sz];
  }
}