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;
long long n, m, k = 0, x, y;
const int P = 239017;
long long t;
int adj[1001][1001], dp[1001] = {}, a[1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> a(n + 1);
  map<int, int> h, c;
  int p = 0;
  set<pair<int, int>> f;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    h[a[i]]++;
    if (h[a[i]] >= 2) {
      f.insert({p + 1, i + 1});
      p = i + 1;
      h = c;
    }
  }
  if (!f.size()) return !(cout << -1);
  cout << f.size() << '\n';
  for (auto r : f)
    cout << r.first << ' ' << (r == *f.rbegin() ? n : r.second) << '\n';
}