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>
const int mx = 1e6 + 5;
const int inf = 0x3f3f3f3f;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; ++i) scanf("%d", &v[i]);
  set<int> s;
  vector<pair<int, int> > pairs;
  int b = 0;
  for (int i = 0; i < n; ++i) {
    if (s.find(v[i]) != s.end()) {
      pairs.push_back({b + 1, i + 1});
      b = i + 1;
      s.clear();
      continue;
    }
    s.insert(v[i]);
  }
  if (!pairs.size()) {
    puts("-1");
    return 0;
  }
  pairs.back().second = n;
  printf("%d\n", int(pairs.size()));
  for (int i = 0; i < pairs.size(); ++i)
    printf("%d %d\n", pairs[i].first, pairs[i].second);
  return 0;
}
