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;
const double EPS = (1e-7);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
bool in_range(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
void solution() {
  int n, x, l = 1;
  scanf("%d", &(n));
  set<int> s;
  vector<pair<int, int> > res;
  for (int i = 0; i < int(n); i++) {
    scanf("%d", &(x));
    if (s.find(x) != s.end()) {
      res.push_back(make_pair(l, i + 1));
      s.clear();
      l = i + 2;
    } else
      s.insert(x);
  }
  if (res.empty())
    puts("-1");
  else {
    printf("%d\n", res.size());
    for (int i = 0; i < int(res.size() - 1); i++)
      printf("%d %d\n", res[i].first, res[i].second);
    printf("%d %d\n", res.back().first, n);
  }
}
int main() {
  double beg = clock();
  solution();
  return 0;
}