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>
using namespace std;
int main() {
  long long int n, i, c = 0, a, x, y, f = 0;
  cin >> n;
  map<long long int, long long int> m1;
  map<long long int, long long int> m2;
  map<long long int, long long int> m3;
  vector<pair<long long int, long long int>> v;
  for (long long int i = 0; i < n; i++) {
    cin >> a;
    m3[a]++;
    if ((m3[a] == 1) && (f == 0)) {
      f = 1;
      x = i + 1;
    } else if (m3[a] == 2) {
      f = 0;
      y = i + 1;
      v.push_back(make_pair(x, y));
      m3.clear();
    }
  }
  if (v.size() == 0) {
    cout << "-1\n";
  } else {
    long long int po = v.size() - 1;
    v[po].second = v[po].second + (n - (v[po].second));
    cout << v.size() << "\n";
    for (i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << "\n";
    }
  }
}
