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 k, n;
struct vec {
  int l;
  int r;
} v[300005];
set<int> s;
int main() {
  cin >> n;
  int x;
  int nr = 0;
  int inc, sf;
  inc = 1;
  sf = 1;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (s.find(x) != s.end()) {
      nr++;
      v[nr].l = inc;
      v[nr].r = sf;
      inc = sf + 1;
      sf++;
      s.clear();
    } else {
      sf++;
      s.insert(x);
    }
  }
  if (nr == 0)
    cout << "-1";
  else {
    sf--;
    if (sf != v[nr].r) v[nr].r = sf;
    cout << nr << "\n";
    for (int i = 1; i <= nr; i++) cout << v[i].l << " " << v[i].r << "\n";
  }
  return 0;
}