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;
const int MAXN = 3 * 1e5 + 5;
int a[MAXN];
map<int, int> cnt;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    if (cnt[a[i]] == ans.size() + 1) {
      cnt[a[i]] = -1;
      ans.push_back(i);
    } else
      cnt[a[i]] = ans.size() + 1;
  }
  if (ans.size() == 0) {
    cout << "-1";
    return 0;
  }
  cout << ans.size() << endl;
  if (ans.back() != n) ans[ans.size() - 1] = n - 1;
  cout << 1 << " " << ans[0] + 1 << endl;
  for (int i = 1; i < ans.size(); ++i) {
    cout << ans[i - 1] + 2 << " " << ans[i] + 1 << endl;
  }
}
