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;
struct node {
  int l, r;
  node(int _l, int _r) : l(_l), r(_r) {}
};
vector<node> ans;
set<int> haha;
int main() {
  int n;
  cin >> n;
  int x;
  int l;
  l = 1;
  scanf("%d", &x);
  haha.insert(x);
  int i = 2;
  while (i <= n) {
    scanf("%d", &x);
    if (haha.find(x) == haha.end())
      haha.insert(x);
    else {
      ans.push_back(node(l, i));
      haha.clear();
      if (i + 1 <= n) {
        scanf("%d", &x);
        l = i + 1;
        haha.insert(x);
        i++;
      }
    }
    i++;
  }
  if (ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  ans[ans.size() - 1].r = n;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].l << " " << ans[i].r << endl;
  return 0;
}