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>
const int MAX = 300001;
using namespace std;
int p[MAX];
int main() {
  std::ios::sync_with_stdio(false);
  set<int> s;
  int n, m, j;
  while (cin >> n) {
    j = 0;
    p[0] = 0;
    for (int i = 1; i < n + 1; i++) {
      cin >> m;
      if (s.count(m)) {
        s.clear();
        p[j++] = i;
      } else
        s.insert(m);
    }
    if (!j)
      cout << "-1" << endl;
    else {
      p[j - 1] = n;
      cout << j << endl;
      cout << "1"
           << " " << p[0] << endl;
      for (int i = 1; i < j; i++) {
        cout << p[i - 1] + 1 << " " << p[i] << endl;
      }
    }
  }
}