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 A[300006];
pair<int, int> Ans[1234567];
int main() {
  ios_base::sync_with_stdio(0);
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
  }
  map<int, int> x;
  int start = 1;
  int cnt = 0;
  for (int i = 1; i <= N; i++) {
    if (x.find(A[i]) != x.end()) {
      Ans[cnt] = make_pair(start, i);
      start = i + 1;
      cnt++;
      x.clear();
    } else {
      x[A[i]]++;
    }
  }
  if (cnt == 0) {
    cout << "-1" << endl;
  } else {
    Ans[cnt - 1].second = N;
    cout << cnt << endl;
    for (int i = 0; i < cnt; i++) {
      cout << Ans[i].first << " " << Ans[i].second << endl;
    }
  }
}
